Atomic and Nuclear Physics Alıştırma Soruları — JAMB UTME — Sayfa 5

Soru 81Soru →

A sample of a radioactive isotope has an initial activity of $8000\text{ Bq}$ . After an elapsed time of $18\text{ minutes}$ , its activity reduces to $1000\text{ Bq}$ . Calculate the decay constant $\lambda$ of the isotope in $\text{min}^{-1}$ .

Cevabı ve açıklamayı göster

Cevap: 0.1155

Cevap

The decay constant of the radioactive isotope is

0.1155\text{ min}^{-1}

.

The activity decreases from

8000\text{ Bq}

to

1000\text{ Bq}

, which is a reduction to

\frac{1}{8}

of its initial value. Since

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

, exactly

3

half-lives have elapsed in

18\text{ minutes}

, meaning

T_{1/2} = 6\text{ minutes}

. Using the relationship

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}}

, we obtain

\lambda \approx 0.1155\text{ min}^{-1}

.

Adım Adım Çözüm

1

Determine the number of elapsed half-lives from the activity reduction ratio.

The fraction of remaining activity is

\frac{A}{A_0} = \frac{1000}{8000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3

, giving

n = 3

half-lives.

Radioactive decay follows the relation

A = A_0 (1/2)^n

.

2

Determine the half-life

T_{1/2}

of the isotope.

T_{1/2} = \frac{t}{n} = \frac{18\text{ min}}{3} = 6\text{ minutes}

.

Total elapsed time is equal to the number of half-lives multiplied by the duration of one half-life.

3

Compute the decay constant

\lambda

in

\text{min}^{-1}

.

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}} = 0.1155\text{ min}^{-1}

.

The decay constant is fundamental to decay rate and related to half-life via

\lambda = \frac{\ln 2}{T_{1/2}}

.

Anahtar Kavram

Radioactive Decay Law and Decay Constant

Tahmini Süre:2m 0s

Soru 82Soru →

If the mass defect of a nucleus is $0.030\text{ u}$ , what is its binding energy? (Take $1\text{ u} = 931\text{ MeV}$ )

27.93\text{ MeV}

27.93\text{ J}

0.030\text{ MeV}

31.033\text{ MeV}

Cevabı ve açıklamayı göster

Cevap: $27.93\text{ MeV}$

Cevap

27.93\text{ MeV}

The binding energy is calculated directly by multiplying the mass defect

\Delta m

by the conversion factor

931\text{ MeV/u}

. Thus,

E_b = 0.030 \times 931 = 27.93\text{ MeV}

.

Adım Adım Çözüm

1

Identify the given values

Mass defect

\Delta m = 0.030\text{ u}

and mass-energy equivalent

1\text{ u} = 931\text{ MeV}

.

To calculate the binding energy, the mass defect in atomic mass units must be converted to energy.

2

Apply mass-energy equivalence conversion

E_b = \Delta m \times 931\text{ MeV/u} = 0.030 \times 931 = 27.93\text{ MeV}

.

Binding energy is equal to mass defect multiplied by the energy equivalent per atomic mass unit.

Anahtar Kavram

Mass Defect and Binding Energy

Soru 83Soru →

During a nuclear reaction, a mass defect of $0.02\text{ u}$ is observed. Given that $1\text{ u} = 931\text{ MeV}$ , what is the total energy released in this reaction?

18.62\text{ MeV}

46\,550\text{ MeV}

931.02\text{ MeV}

18.62\text{ J}

Cevabı ve açıklamayı göster

Cevap: $18.62\text{ MeV}$

Cevap

18.62\text{ MeV}

The energy released in a nuclear reaction is determined by multiplying the mass defect

\Delta m

by the energy equivalence factor (

931\text{ MeV}

per atomic mass unit). Multiplying

0.02\text{ u}

by

931\text{ MeV/u}

yields

18.62\text{ MeV}

.

Adım Adım Çözüm

1

Identify the mass defect and the energy equivalence factor.

Mass defect

\Delta m = 0.02\text{ u}

and

1\text{ u} = 931\text{ MeV}

.

The energy released in a nuclear reaction is directly proportional to its mass defect.

2

Calculate the total energy released by multiplying mass defect by energy equivalent per unit mass.

E = 0.02\text{ u} \times 931\text{ MeV/u} = 18.62\text{ MeV}

.

Applying the conversion factor yields the energy released in

\text{MeV}

.

Anahtar Kavram

Mass defect and energy conversion in nuclear reactions

Soru 84Soru →

A Geiger-Müller counter records a total count rate of $340\text{ counts per minute}$ near a radioactive source. The background radiation in the laboratory produces a steady count rate of $20\text{ counts per minute}$ . If the total count rate recorded by the counter drops to $60\text{ counts per minute}$ after an elapsed time of $15\text{ minutes}$ , what is the half-life of the radioactive source in minutes?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The half-life of the radioactive source is

5\text{ minutes}

.

To find the true activity of the radioactive source, the constant background radiation of

20\text{ cpm}

must be subtracted from all detector readings. The initial source activity is

340 - 20 = 320\text{ cpm}

and the activity after

15\text{ minutes}

is

60 - 20 = 40\text{ cpm}

. The fraction of source activity remaining is

40 / 320 = 1/8 = (1/2)^3

, which means

3

half-lives have elapsed in

15\text{ minutes}

. Dividing the total time by the number of half-lives (

15 / 3

) gives a half-life of

5\text{ minutes}

.

Adım Adım Çözüm

1

Calculate the initial activity of the radioactive source by subtracting the background count rate.

A_0 = 340\text{ cpm} - 20\text{ cpm} = 320\text{ cpm}

Background radiation contributes to the detector reading and must be isolated from the source activity.

2

Calculate the activity of the source after 15 minutes by subtracting the background count rate.

A(t) = 60\text{ cpm} - 20\text{ cpm} = 40\text{ cpm}

The background count remains constant at

20\text{ cpm}

, so the actual count due to the source is

40\text{ cpm}

.

3

Determine the remaining fraction of the radioactive source.

\frac{A(t)}{A_0} = \frac{40}{320} = \frac{1}{8}

Radioactive decay follows an exponential decay law based on the fraction of initial undecayed nuclei.

4

Calculate the number of elapsed half-lives

n

.

\left(\frac{1}{2}\right)^n = \frac{1}{8} = \left(\frac{1}{2}\right)^3 \implies n = 3

The remaining fraction equals

(1/2)^n

where

n

is the number of half-lives.

5

Compute the half-life

T_{1/2}

.

T_{1/2} = \frac{t}{n} = \frac{15\text{ minutes}}{3} = 5\text{ minutes}

The total elapsed time is the product of the number of half-lives and the duration of one half-life.

Anahtar Kavram

Radioactive Decay Law and Half-life with Background Radiation Correction

Soru 85Soru →

A sample of a radioactive substance has a decay constant of $0.0154\text{ h}^{-1}$ . What is the elapsed time, in hours, required for $87.5\%$ of the original sample to decay? (Take $\ln 2 = 0.693$ )

Cevabı ve açıklamayı göster

Cevap: 135

Cevap

135 hours

First, calculate the half-life of the substance using the relation

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.0154\text{ h}^{-1}} = 45\text{ hours}

. Since

87.5\%

of the sample has decayed, the remaining fraction of the sample is

100\% - 87.5\% = 12.5\% = \frac{1}{8}

. Expressing

\frac{1}{8}

as a power of

\frac{1}{2}

gives

\left(\frac{1}{2}\right)^3

, indicating that

3

half-lives have passed. The total elapsed time is therefore

3 \times 45\text{ hours} = 135\text{ hours}

.

Adım Adım Çözüm

1

Calculate the half-life from the given decay constant

Half-life

T_{1/2} = 45\text{ hours}

The decay constant

\lambda

and half-life

T_{1/2}

are related by the formula

T_{1/2} = \frac{\ln 2}{\lambda}

2

Find the remaining percentage and fraction of the sample

Remaining fraction is

12.5\%

or

\frac{1}{8}

Radioactive decay equations use the undecayed remaining amount, which is

100\% - 87.5\% = 12.5\%

3

Calculate the number of half-lives elapsed

Number of half-lives

n = 3

Since

\left(\frac{1}{2}\right)^n = \frac{1}{8}

, solving for

n

gives

n = 3

4

Compute the total elapsed time

Total elapsed time

t = 135\text{ hours}

Total elapsed time is the product of the number of half-lives and the half-life duration (

3 \times 45\text{ hours}

)

Anahtar Kavram

Radioactive Decay Law and Half-life Relationship

Soru 86Soru →

A sample of a radioactive nuclide has a half-life of $4\text{ hours}$ . What is the ratio of the number of nuclei that have decayed to the number of undecayed nuclei remaining after a total elapsed time of $16\text{ hours}$ ?

15 : 1

1 : 15

15 : 16

7 : 1

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Cevap: $15 : 1$

Cevap

The ratio of decayed nuclei to remaining undecayed nuclei is

15 : 1

.

In an elapsed time of

16\text{ hours}

(which is

4

half-lives), the remaining undecayed portion of the sample is

(1/2)^4 = 1/16

of the original amount. Consequently,

1 - 1/16 = 15/16

of the nuclei have decayed. The ratio of decayed nuclei to remaining nuclei is

(15/16) : (1/16) = 15 : 1

.

Adım Adım Çözüm

1

Determine the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{16\text{ hours}}{4\text{ hours}} = 4\text{ half-lives}

.

The number of half-lives is the total time divided by the half-life duration.

2

Calculate the fraction of original nuclei remaining undecayed (

N/N_0

).

\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^4 = \frac{1}{16}

.

After

n

half-lives, the remaining fraction is

(1/2)^n

.

3

Find the fraction of original nuclei that have decayed (

N_{\text{decayed}}/N_0

).

\frac{N_{\text{decayed}}}{N_0} = 1 - \frac{N}{N_0} = 1 - \frac{1}{16} = \frac{15}{16}.

Decayed fraction equals total initial fraction (1) minus the undecayed fraction.

4

Compute the ratio of decayed nuclei to remaining undecayed nuclei.

\text{Ratio} = \frac{N_{\text{decayed}}}{N} = \frac{15/16}{1/16} = 15 : 1.

The question asks for the ratio of decayed nuclei to remaining undecayed nuclei.

Anahtar Kavram

Radioactive Decay Law and Half-life calculations involving ratios of decayed versus remaining quantities.

Soru 87Soru →

A Plutonium-239 nucleus (

{^{239}_{94}\text{Pu}}

) captures a thermal neutron (

{^{1}_{0}\text{n}}

) and undergoes nuclear fission according to the reaction equation:

{^{239}_{94}\text{Pu}} + {^{1}_{0}\text{n}} \rightarrow {^{144}_{56}\text{Ba}} + {^{93}_{38}\text{Sr}} + x\,{^{1}_{0}\text{n}} + Q

Given the rest masses:
- Mass of ${^{239}_{94}\text{Pu}} = 239.0522\text{ u}$
- Mass of ${^{1}_{0}\text{n}} = 1.0087\text{ u}$
- Mass of ${^{144}_{56}\text{Ba}} = 143.9229\text{ u}$
- Mass of ${^{93}_{38}\text{Sr}} = 92.9154\text{ u}$
- $1\text{ u} = 931.5\text{ MeV}$

What is the number of emitted neutrons $x$ and the total energy $Q$ released in this fission process?

x = 3

neutrons and

Q = 183.04\text{ MeV}

x = 3

neutrons and

Q = 756.56\text{ MeV}

x = 2

neutrons and

Q = 183.04\text{ MeV}

x = 3

neutrons and

Q = 2.11 \times 10^{-4}\text{ MeV}

Cevabı ve açıklamayı göster

Cevap: $x = 3$ neutrons and $Q = 183.04\text{ MeV}$

Cevap

The reaction releases

x = 3

neutrons and an energy

Q = 183.04\text{ MeV}

.

The conservation of nucleon number requires

239 + 1 = 144 + 93 + x

, giving

x = 3

neutrons. The mass defect is calculated by subtracting the mass of products (

143.9229 + 92.9154 + 3 \times 1.0087 = 239.8644\text{ u}

) from the mass of reactants (

239.0522 + 1.0087 = 240.0609\text{ u}

), yielding

\Delta m = 0.1965\text{ u}

. Converting this to energy gives

Q = 0.1965 \times 931.5\text{ MeV} = 183.04\text{ MeV}

.

Adım Adım Çözüm

1

Balance mass number (

A

) to find the number of neutrons

x

239 + 1 = 144 + 93 + x \implies 240 = 237 + x \implies x = 3

Total mass number must be conserved in a nuclear reaction.

2

Calculate the total mass of the reactants (

m_{\text{reactants}}

)

m_{\text{reactants}} = 239.0522\text{ u} + 1.0087\text{ u} = 240.0609\text{ u}

The reactants consist of one Plutonium-239 nucleus and one thermal neutron.

3

Calculate the total mass of the products (

m_{\text{products}}

)

m_{\text{products}} = 143.9229\text{ u} + 92.9154\text{ u} + 3(1.0087\text{ u}) = 239.8644\text{ u}

The products consist of one Barium-144 nucleus, one Strontium-93 nucleus, and three neutrons.

4

Calculate the mass defect (

\Delta m

)

\Delta m = 240.0609\text{ u} - 239.8644\text{ u} = 0.1965\text{ u}

Mass defect is the difference between total mass of reactants and total mass of products.

5

Calculate the liberated energy (

Q

)

Q = 0.1965\text{ u} \times 931.5\text{ MeV/u} = 183.03975\text{ MeV} \approx 183.04\text{ MeV}

Using Einstein's mass-energy equivalence conversion factor

1\text{ u} = 931.5\text{ MeV}

.

Anahtar Kavram

Nuclear fission conservation laws (mass/atomic number conservation) and mass defect energy equivalence (

Q = \Delta m \cdot 931.5\text{ MeV/u}

).

Tahmini Süre:2m 0s

Soru 88Soru →

A subatomic particle of mass $6.63 \times 10^{-31}\text{ kg}$ moves with a velocity of $1.0 \times 10^6\text{ m/s}$ . Calculate its de Broglie wavelength in nanometers ( $\text{nm}$ ). (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ and $1\text{ nm} = 10^{-9}\text{ m}$ )

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The de Broglie wavelength of the particle is

1.0\text{ nm}

.

Using de Broglie's wave-particle duality relation

\lambda = \frac{h}{p} = \frac{h}{m v}

, substituting the given values yields

\lambda = \frac{6.63 \times 10^{-34}\text{ J}\cdot\text{s}}{(6.63 \times 10^{-31}\text{ kg}) \times (1.0 \times 10^6\text{ m/s})} = 1.0 \times 10^{-9}\text{ m}

. In nanometers, this is equal to

1.0\text{ nm}

.

Adım Adım Çözüm

1

Calculate the linear momentum of the particle

p = 6.63 \times 10^{-25}\text{ kg}\cdot\text{m/s}

Momentum is the product of mass and velocity (

p = m v

).

2

Calculate the de Broglie wavelength

\lambda = 1.0 \times 10^{-9}\text{ m}

According to de Broglie's hypothesis, wavelength is given by

\lambda = \frac{h}{p}

.

3

Convert the calculated wavelength to nanometers

\lambda = 1.0\text{ nm}

Since

1\text{ nm} = 10^{-9}\text{ m}

, dividing

1.0 \times 10^{-9}\text{ m}

by

10^{-9}

yields

1.0\text{ nm}

.

Anahtar Kavram

de Broglie Wavelength and Wave-Particle Duality

Soru 89Soru →

For a given clean photosensitive metal surface, doubling the intensity of incident monochromatic radiation of frequency $f$ (where $f > f_0$ ) doubles the stopping potential required to reduce the photoelectric current to zero.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is False. Stopping potential depends exclusively on the frequency of the incident radiation and the work function of the metal emitter, remaining independent of light intensity.

The statement is False because stopping potential

V_s

is defined by

e V_s = hf - W_0

. Doubling the intensity of the incident radiation increases the number of emitted photoelectrons per second (photocurrent) but does not alter the energy per photon

hf

or the maximum kinetic energy of the photoelectrons. Therefore, the stopping potential needed to halt the current remains unchanged.

Adım Adım Çözüm

1

Analyze the physical meaning of light intensity in quantum theory.

Increasing the light intensity increases the photon flux (number of photons per unit area per second), thereby increasing the rate of electron emission (photocurrent).

Intensity corresponds to the rate of photon arrival, not the energy of individual photons.

2

Relate photon energy and work function to stopping potential using Einstein's photoelectric equation.

e V_s = K_{\text{max}} = hf - W_0

, where

V_s

is the stopping potential,

f

is frequency, and

W_0

is work function.

Each emitted electron absorbs energy from a single photon.

3

Evaluate the effect of doubling radiation intensity on stopping potential.

Because

f

and

W_0

are constant,

K_{\text{max}}

and

V_s

remain unaltered.

Stopping potential is governed by photon frequency rather than total light intensity.

Anahtar Kavram

Independence of photoelectron maximum kinetic energy and stopping potential from incident light intensity

Soru 90Soru →

A nucleus contains $6$ protons and $6$ neutrons. Given that the mass of a proton is $1.0078\text{ u}$ , the mass of a neutron is $1.0087\text{ u}$ , and the total binding energy of the nucleus is $93.10\text{ MeV}$ , what is the mass of the nucleus in atomic mass units ( $\text{u}$ )? (Take $1\text{ u} = 931\text{ MeV}$ )

Cevabı ve açıklamayı göster

Cevap: 11.999

Cevap

The mass of the nucleus is

11.9990\text{ u}

.

To find the mass of the nucleus, first calculate the combined mass of the 6 free protons and 6 free neutrons:

6(1.0078\text{ u}) + 6(1.0087\text{ u}) = 12.0990\text{ u}

. Next, determine the mass defect from the binding energy:

\Delta m = \frac{93.10\text{ MeV}}{931\text{ MeV/u}} = 0.1000\text{ u}

. Finally, subtract the mass defect from the total mass of individual nucleons to get the nuclear mass:

12.0990\text{ u} - 0.1000\text{ u} = 11.9990\text{ u}

.

Adım Adım Çözüm

1

Calculate the total mass of the individual free protons and neutrons.

Total mass of nucleons =

6(1.0078\text{ u}) + 6(1.0087\text{ u}) = 12.0990\text{ u}

.

The mass of the constituent particles before binding is the sum of their individual rest masses.

2

Calculate the mass defect from the given binding energy.

Mass defect

\Delta m = \frac{93.10\text{ MeV}}{931\text{ MeV/u}} = 0.1000\text{ u}

.

Binding energy and mass defect are related by

E_b = \Delta m \times 931\text{ MeV/u}

.

3

Subtract the mass defect from the total nucleon mass to obtain the nuclear mass.

Mass of nucleus =

12.0990\text{ u} - 0.1000\text{ u} = 11.9990\text{ u}

.

Mass defect represents the mass lost when nucleons bind together into a nucleus.

Anahtar Kavram

Mass Defect and Binding Energy Relationship

Soru 91Soru →

During natural radioactive decay, an unstable nucleus can emit a beta-minus ( $\beta^-$ ) particle despite the nucleus containing no free electrons. Which of the following mechanisms correctly describes the origin of this emitted particle?

A neutron inside the nucleus transforms into a proton, emitting an electron and an antineutrino.

An orbital electron from the innermost atomic K-shell is ejected due to intense nuclear energy.

A proton inside the nucleus converts into a neutron, releasing a negatively charged electron.

High-energy gamma radiation within the nucleus spontaneously converts into an electron-positron pair.

Cevabı ve açıklamayı göster

Cevap: A neutron inside the nucleus transforms into a proton, emitting an electron and an antineutrino.

Cevap

A neutron inside the nucleus transforms into a proton, emitting an electron and an antineutrino.

The emission of a beta-minus particle results from a weak interaction process within an unstable nucleus where a neutron converts into a proton, emitting a high-speed electron (beta particle) and an antineutrino. This preserves total electric charge (

0 = +1 - 1

) and nucleon number (

1 = 1 + 0

).

Adım Adım Çözüm

1

Identify the constituents of the nucleus and the nature of the emitted particle.

The nucleus consists of protons and neutrons, while a beta-minus (

\beta^-

) particle is a fast-moving electron (

^{0}_{-1}e

).

Since electrons do not exist as independent bound particles inside the nucleus, the emitted electron must be created during a nuclear transformation.

2

Apply conservation laws (charge and mass number) to determine the nuclear reaction.

^{1}_{0}n \rightarrow ^{1}_{1}p + ^{0}_{-1}e + \bar{\nu}_e

A neutron (

^{1}_{0}n

) decays into a proton (

^{1}_{1}p

), producing a beta particle (

^{0}_{-1}e

) and an antineutrino (

\bar{\nu}_e

) to conserve atomic number, mass number, and lepton number.

Anahtar Kavram

Mechanism of Nuclear Beta Decay

Soru 92Soru →

An electron in a hydrogen atom makes a transition from the third energy level ( $n = 3$ ) to the second energy level ( $n = 2$ ). If the energy of the electron at $n = 3$ is $-1.51\text{ eV}$ and at $n = 2$ is $-3.40\text{ eV}$ , what is the frequency of the emitted photon? ( $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

4.58 \times 10^{14}\text{ Hz}

2.86 \times 10^{33}\text{ Hz}

1.19 \times 10^{15}\text{ Hz}

8.24 \times 10^{14}\text{ Hz}

Cevabı ve açıklamayı göster

Cevap: $4.58 \times 10^{14}\text{ Hz}$

Cevap

The frequency of the emitted photon is

4.58 \times 10^{14}\text{ Hz}

.

When an electron transitions from a higher energy state to a lower energy state, it emits a photon whose energy equals the difference between the two levels:

\Delta E = E_3 - E_2 = 1.89\text{ eV}

. Converting

1.89\text{ eV}

to Joules gives

3.024 \times 10^{-19}\text{ J}

. Dividing by Planck's constant (

6.6 \times 10^{-34}\text{ J}\cdot\text{s}

) gives a frequency of

4.58 \times 10^{14}\text{ Hz}

.

Adım Adım Çözüm

1

Calculate the energy difference between the initial and final energy levels.

\Delta E = E_3 - E_2 = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

The energy of the emitted photon equals the difference in energy between the two states.

2

Convert the energy change from electron-volts (eV) to Joules (J).

\Delta E = 1.89 \times 1.6 \times 10^{-19}\text{ J} = 3.024 \times 10^{-19}\text{ J}

SI units (Joules) are required to calculate frequency using Planck's constant in

\text{J}\cdot\text{s}

.

3

Apply the photon energy formula

E = hf

to find the frequency

f

.

f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J}\cdot\text{s}} = 4.5818 \times 10^{14}\text{ Hz} \approx 4.58 \times 10^{14}\text{ Hz}

Dividing energy by Planck's constant yields the photon frequency.

Anahtar Kavram

Photon Emission and Energy Level Transition

Soru 93Soru →

In a nuclear fusion process, two deuterium nuclei (

\text{^{2}_{1}H}

) fuse to form a helium-3 nucleus (

\text{^{3}_{2}He}

) and a neutron (

\text{^{1}_{0}n}

) according to the reaction equation:

\text{^{2}_{1}H} + \text{^{2}_{1}H} \rightarrow \text{^{3}_{2}He} + \text{^{1}_{0}n} + Q

Given the mass values:
- Mass of $\text{^{2}_{1}H} = 2.0141\text{ u}$
- Mass of $\text{^{3}_{2}He} = 3.0160\text{ u}$
- Mass of $\text{^{1}_{0}n} = 1.0087\text{ u}$

Using the conversion factor $1\text{ u} = 931.5\text{ MeV}$ , what is the energy released ( $Q$ ) in this fusion reaction in $\text{MeV}$ ?

Cevabı ve açıklamayı göster

Cevap: 3.26

Cevap

The total energy released (

Q

) in the reaction is

3.26\text{ MeV}

.

The energy released in a nuclear fusion reaction is proportional to the decrease in total rest mass (mass defect). Summing the mass of two deuterium nuclei gives

4.0282\text{ u}

, while the sum of the product masses (

\text{^{3}_{2}He}

and a neutron) is

4.0247\text{ u}

. Subtracting these yields a mass defect of

0.0035\text{ u}

. Multiplying

0.0035\text{ u}

by

931.5\text{ MeV/u}

gives

3.26\text{ MeV}

of released energy.

Adım Adım Çözüm

1

Calculate the total initial mass of the reacting deuterium nuclei.

m_{\text{reactants}} = 2 \times 2.0141\text{ u} = 4.0282\text{ u}

Two deuterium nuclei participate on the reactant side of the equation.

2

Calculate the total final mass of the products.

m_{\text{products}} = 3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

The reaction produces one helium-3 nucleus and one neutron.

3

Determine the mass defect (difference between reactant and product masses).

\Delta m = 4.0282\text{ u} - 4.0247\text{ u} = 0.0035\text{ u}

The mass lost during fusion is converted into nuclear kinetic energy and radiation.

4

Convert the mass defect into energy in MeV using the conversion factor.

Q = 0.0035\text{ u} \times 931.5\text{ MeV/u} = 3.26025\text{ MeV} \approx 3.26\text{ MeV}

Each atomic mass unit lost corresponds to

931.5\text{ MeV}

of energy.

Anahtar Kavram

Mass defect and energy release in nuclear fusion reactions (

E = \Delta m c^2

)

Soru 94Soru →

In a nuclear fusion reaction, two deuterium nuclei ( ${^{2}_{1}\text{H}}$ ) combine to form a helium-3 nucleus ( ${^{3}_{2}\text{He}}$ ) and a neutron ( ${^{1}_{0}\text{n}}$ ). The total mass of the two reactant deuterium nuclei is $4.0282\text{ u}$ , while the total mass of the resulting helium-3 and neutron products is $4.0247\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , calculate the total energy released in this reaction in $\text{MeV}$ .

Cevabı ve açıklamayı göster

Cevap: 3.26

Cevap

The total energy released in the nuclear fusion reaction is approximately 3.26 MeV.

The energy released in a nuclear fusion reaction is determined by the mass defect, which is the difference between the total mass of reactants and the total mass of products. Subtracting

4.0247\text{ u}

from

4.0282\text{ u}

yields a mass defect of

0.0035\text{ u}

. Multiplying this mass defect by the mass-energy conversion factor of

931.5\text{ MeV/u}

gives an energy release of approximately

3.26\text{ MeV}

.

Adım Adım Çözüm

1

Calculate the mass defect (Δm)

Δm = 4.0282 u - 4.0247 u = 0.0035 u

Mass defect is the difference between the initial mass of reactants and the final mass of products in a nuclear reaction.

2

Calculate the energy released (E) in MeV

E = 0.0035 u × 931.5 MeV/u = 3.26025 MeV

According to mass-energy equivalence, 1 unified atomic mass unit (u) liberates 931.5 MeV of energy.

Anahtar Kavram

Mass defect and energy release in nuclear fusion

Soru 95Soru →

If the kinetic energy of a non-relativistic electron is increased by a factor of $4$ , how does its associated de Broglie wavelength change?

It decreases to one-half of its original value

It decreases to one-fourth of its original value

It increases by a factor of

2

It increases by a factor of

4

Cevabı ve açıklamayı göster

Cevap: It decreases to one-half of its original value

Cevap

The de Broglie wavelength decreases to one-half of its original value.

The de Broglie wavelength

\lambda

of a particle is given by

\lambda = \frac{h}{p}

, where momentum

p = \sqrt{2m E_k}

. Substituting momentum into the wavelength equation yields

\lambda = \frac{h}{\sqrt{2m E_k}}

. If kinetic energy

E_k

quadruples, the denominator increases by a factor of

\sqrt{4} = 2

, which reduces the wavelength to half of its initial value.

Adım Adım Çözüm

1

Relate de Broglie wavelength to momentum and kinetic energy.

The de Broglie wavelength is given by

\lambda = \frac{h}{p}

. Since kinetic energy

E_k = \frac{p^2}{2m}

, momentum is

p = \sqrt{2m E_k}

. Thus,

\lambda = \frac{h}{\sqrt{2m E_k}}

.

Establishing the mathematical relationship between wavelength

\lambda

and kinetic energy

E_k

.

2

Apply the scaling factor of 4 to the kinetic energy.

When

E_k' = 4 E_k

, the new wavelength

\lambda'

is

\lambda' = \frac{h}{\sqrt{2m (4 E_k)}} = \frac{h}{2\sqrt{2m E_k}} = \frac{\lambda}{2}

.

Evaluating the square root factor

\sqrt{4} = 2

in the denominator.

3

Conclude the final ratio.

The new wavelength is half the original wavelength.

The de Broglie wavelength is inversely proportional to the square root of kinetic energy.

Anahtar Kavram

Relationship between de Broglie wavelength and kinetic energy

Soru 96Soru →

Consider the nuclear fission reaction represented by the equation below:

{^{235}_{92}\text{U}} + {^{1}_{0}\text{n}} \rightarrow {^{144}_{56}\text{Ba}} + {^{89}_{36}\text{Kr}} + x\,^{1}_{0}\text{n}

What is the value of

x

, representing the number of neutrons emitted in this reaction?

1

2

3

4

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The number of emitted neutrons,

x

, is equal to 3.

By the law of conservation of mass number, the sum of nucleon numbers on the left side (

235 + 1 = 236

) must equal the sum on the right side (

144 + 89 + x

). Solving the equation

236 = 233 + x

gives

x = 3

.

Adım Adım Çözüm

1

Calculate the total mass number on the reactant side.

Total reactant mass number =

235 + 1 = 236

.

According to the law of conservation of nucleon number, the total mass number before the reaction must equal the total mass number after the reaction.

2

Sum the known mass numbers on the product side.

Known product mass number =

144 + 89 = 233

.

Adding the mass numbers of Barium-144 and Krypton-89.

3

Set up and solve the mass conservation equation for

x

.

236 = 233 + x(1) \implies x = 3

.

Each neutron has a mass number of 1, so solving

236 - 233

gives

x = 3

.

Anahtar Kavram

Conservation of Mass Number in Nuclear Fission

Soru 97Soru →

How much energy, in $\text{MeV}$ , is released when a nuclear fission process results in a mass defect of $0.05 \text{ u}$ ? (Take $1 \text{ u} = 931.5 \text{ MeV}$ )

Cevabı ve açıklamayı göster

Cevap: 46.575

Cevap

46.575 MeV

The total energy released in a nuclear fission reaction is found by multiplying the mass defect by the energy equivalent of 1 atomic mass unit. Multiplying

0.05 \text{ u}

by

931.5 \text{ MeV/u}

gives

46.575 \text{ MeV}

.

Adım Adım Çözüm

1

Identify the mass defect and the mass-to-energy conversion factor.

Mass defect

\Delta m = 0.05 \text{ u}

, and

1 \text{ u} = 931.5 \text{ MeV}

.

In nuclear fission, mass lost during the reaction is converted directly into energy according to Einstein's mass-energy equivalence principle.

2

Calculate the total energy released.

E = 0.05 \text{ u} \times 931.5 \text{ MeV/u} = 46.575 \text{ MeV}

.

Multiplying the mass defect in atomic mass units by

931.5 \text{ MeV/u}

gives the total released energy in

\text{MeV}

.

Anahtar Kavram

Mass-Energy Conversion in Nuclear Reactions

Soru 98Soru →

In a photoelectric cell experiment, monochromatic light of frequency $f$ and intensity $I$ illuminates a potassium surface, emitting photoelectrons with a stopping potential of $1.5\text{ V}$ . When the light frequency is increased to $1.5f$ and the intensity is set to $2I$ , the stopping potential rises to $3.5\text{ V}$ . What is the stopping potential if the light intensity is further increased to $4I$ while maintaining the frequency constant at $1.5f$ ?

3.5\text{ V}

7.0\text{ V}

6.0\text{ V}

14.0\text{ V}

Cevabı ve açıklamayı göster

Cevap: $3.5\text{ V}$

Cevap

3.5\text{ V}

According to Einstein's photoelectric theory, stopping potential

V_s

is directly proportional to maximum kinetic energy (

e V_s = hf - W_0

). It depends exclusively on the frequency of incident radiation and the target work function. Changing light intensity from

2I

to

4I

increases the number of emitted photoelectrons per second, but does not alter maximum kinetic energy. Therefore, at frequency

1.5f

, the stopping potential remains

3.5\text{ V}

.

Adım Adım Çözüm

1

Identify the relationship between stopping potential, frequency, and light intensity in the photoelectric effect.

Einstein's photoelectric equation states that

e V_s = K_{\text{max}} = h f - W_0

, where

V_s

is stopping potential,

f

is frequency, and

W_0

is work function.

Stopping potential measures the maximum kinetic energy of emitted photoelectrons.

2

Analyze the impact of changing light intensity while holding frequency constant at

1.5f

.

Increasing intensity from

2I

to

4I

increases the number of incident photons per second (hence increasing emission current), but individual photon energy

E = hf

remains unchanged.

Photon energy depends solely on frequency

f

, not on radiation intensity.

3

Determine the new stopping potential value.

Since the frequency remains fixed at

1.5f

,

V_s

stays at

3.5\text{ V}

.

Maximum kinetic energy and stopping potential are independent of light intensity.

Anahtar Kavram

Independence of photoelectron kinetic energy and stopping potential from light intensity

Soru 99Soru →

A Uranium-235 nucleus undergoes nuclear fission after absorbing a thermal neutron according to the reaction equation:

{^{235}_{92}\text{U}} + {^{1}_{0}\text{n}} \rightarrow {^{140}_{54}\text{Xe}} + {^{94}_{38}\text{Sr}} + x\ {^{1}_{0}\text{n}}

Given the rest masses:
- Mass of

{^{235}_{92}\text{U}} = 235.0439\text{ u}

- Mass of

{^{140}_{54}\text{Xe}} = 139.9216\text{ u}

- Mass of

{^{94}_{38}\text{Sr}} = 93.9154\text{ u}

- Mass of

{^{1}_{0}\text{n}} = 1.0087\text{ u}

-

1\text{ u} = 931.5\text{ MeV}

Calculate the total energy released during this fission process in MeV.

Cevabı ve açıklamayı göster

Cevap: 184.62

Cevap

184.62 MeV

The total energy released is calculated by first balancing the nuclear equation to determine that 2 neutrons are emitted (

x = 2

). The total mass of reactants is

236.0526\text{ u}

and the products is

235.8544\text{ u}

. The mass defect of

0.1982\text{ u}

multiplied by

931.5\text{ MeV/u}

yields

184.62\text{ MeV}

.

Adım Adım Çözüm

1

Balance the mass numbers in the nuclear equation to find the number of emitted neutrons (

x

)

x = 2

neutrons

Conservation of mass number requires

235 + 1 = 140 + 94 + x

, giving

236 = 234 + x

, so

x = 2

.

2

Calculate the total mass of the reactants before fission

236.0526\text{ u}

Summing the mass of U-235 (

235.0439\text{ u}

) and the incident neutron (

1.0087\text{ u}

).

3

Calculate the total mass of the products after fission

235.8544\text{ u}

Summing masses of Xe-140 (

139.9216\text{ u}

), Sr-94 (

93.9154\text{ u}

), and 2 neutrons (

2 \times 1.0087\text{ u} = 2.0174\text{ u}

).

4

Determine the mass defect (mass difference)

\Delta m = 0.1982\text{ u}

Mass defect

\Delta m = m_{\text{reactants}} - m_{\text{products}} = 236.0526 - 235.8544 = 0.1982\text{ u}

.

5

Convert mass defect to energy using

1\text{ u} = 931.5\text{ MeV}

184.62\text{ MeV}

Multiplying mass defect

0.1982\text{ u}

by

931.5\text{ MeV/u}

gives

184.6233\text{ MeV}

, which rounds to

184.62\text{ MeV}

.

Anahtar Kavram

Mass defect and mass-energy equivalence in nuclear fission reactions

Tahmini Süre:2m 30s

Soru 100Soru →

An $\alpha$ -particle (charge $+2e$ , mass $m_\alpha$ ) and a $\beta^-$ -particle (charge $-e$ , mass $m_\beta$ ) emitted during natural radioactive decay enter a region containing uniform, mutually perpendicular electric ( $E$ ) and magnetic ( $B$ ) fields. Both particles move along paths perpendicular to both fields and traverse the region without undergoing any deflection. What is the ratio of the kinetic energy of the $\alpha$ -particle to that of the $\beta^-$ -particle, $\frac{E_{k,\alpha}}{E_{k,\beta}}$ ?

\frac{m_\alpha}{m_\beta}

1

2

\frac{m_\alpha}{2m_\beta}

Cevabı ve açıklamayı göster

Cevap: $\frac{m_\alpha}{m_\beta}$

Cevap

The ratio of their kinetic energies is equal to the ratio of their masses,

\frac{m_\alpha}{m_\beta}

.

In crossed uniform electric and magnetic fields acting as a velocity selector, a charged particle passes undeflected when the electric force

qE

balances the magnetic Lorentz force

qvB

. Equating these forces yields

v = \frac{E}{B}

, which depends only on field strengths and is independent of mass and charge. Because both the

\alpha

-particle and

\beta^-

-particle traverse undeflected, both possess the same speed

v

. Substituting equal speeds into the kinetic energy formula

E_k = \frac{1}{2}mv^2

leaves the ratio of their kinetic energies equal to the ratio of their rest masses,

\frac{m_\alpha}{m_\beta}

.

Adım Adım Çözüm

1

Apply the force balance condition for particles moving undeflected in crossed electric and magnetic fields.

Electric force magnitude

F_E = qE

must equal magnetic force magnitude

F_B = qvB

, giving

qE = qvB \implies v = \frac{E}{B}

.

For zero net deflection, the electrostatic force and magnetic Lorentz force must be equal in magnitude and opposite in direction.

2

Determine the velocities of the

\alpha

-particle and

\beta^-

-particle.

v_\alpha = \frac{E}{B}

and

v_\beta = \frac{E}{B}

, so

v_\alpha = v_\beta = v

.

The velocity selection equation

v = \frac{E}{B}

is completely independent of particle mass

m

and charge

q

.

3

Express the ratio of the kinetic energies of the two emissions using

E_k = \frac{1}{2}mv^2

.

\frac{E_{k,\alpha}}{E_{k,\beta}} = \frac{\frac{1}{2} m_\alpha v^2}{\frac{1}{2} m_\beta v^2} = \frac{m_\alpha}{m_\beta}

.

Since the speed

v

is identical for both particles, the

\frac{1}{2}v^2

terms cancel out completely.

Anahtar Kavram

Velocity selector behavior and kinetic energy dependence of radiation emissions in electromagnetic fields

Atomic and Nuclear Physics

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