Atomic and Nuclear Physics Practice Questions — JAMB UTME — Page 6

Which of the following types of radiation emitted during natural radioactive decay has the greatest ionizing power?

\alpha

-particles

\beta

-particles

\gamma

-rays

X-rays

Show answer & explanation

Answer: $\alpha$ -particles

Answer

Alpha particles (

\alpha

-particles) have the highest ionizing power among natural radioactive emissions.

Alpha particles have the largest charge (

+2e

) and mass among natural nuclear emissions, allowing them to exert strong electrostatic forces on atoms in their path and strip away electrons easily, giving them the highest ionizing capability.

Step-by-Step Solution

1

Identify the nature, mass, and charge of natural radiation emissions

\alpha

-particles consist of helium nuclei (

^4_2\text{He}

) with mass

4\text{ u}

and charge

+2e

;

\beta

-particles are high-speed electrons with negligible mass and charge

-e

;

\gamma

-rays are neutral photons.

Ionization depends directly on charge magnitude and kinetic energy transfer capacity during interactions with matter.

2

Compare the ionizing power of the emissions

Because

\alpha

-particles carry a large

+2e

charge and move relatively slowly due to their larger mass, they interact strongly with orbital electrons, knocking them off atoms easily along a short path.

Relative ionizing power follows the ratio of roughly

10,000 : 100 : 1

for

\alpha : \beta : \gamma

emissions.

Key Concept

Ionizing power vs. Penetrating power of radioactive emissions

Estimated Time:45s

Question 102Question →

The mass defect of a nitrogen nucleus $^{14}_{7}\text{N}$ is calculated to be $0.112\text{ u}$ . Taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ?

Show answer & explanation

Answer: 104.328

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

104.328\text{ MeV}

.

The total binding energy of a nucleus is equal to the mass defect multiplied by the energy equivalent of one atomic mass unit (

931.5\text{ MeV/u}

). Calculating

0.112\text{ u} \times 931.5\text{ MeV/u}

gives

104.328\text{ MeV}

.

Step-by-Step Solution

1

Apply the binding energy formula

E_b = \Delta m \times 931.5\text{ MeV/u}

.

E_b = 0.112\text{ u} \times 931.5\text{ MeV/u} = 104.328\text{ MeV}

.

The binding energy is obtained by converting the mass defect directly into its energy equivalent.

Key Concept

Mass Defect and Binding Energy Conversion

Question 103Question →

A nitrogen nucleus $^{14}_{7}\text{N}$ has a nuclear mass of $13.9992\text{ u}$ . Given that the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ? (Take $1\text{ u} = 931\text{ MeV}$ )

Show answer & explanation

Answer: 105.02

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

105.02\text{ MeV}

.

The total binding energy is computed by finding the total rest mass of 7 protons and 7 neutrons (14.1120 u), subtracting the actual nuclear mass of Nitrogen-14 (13.9992 u) to obtain a mass defect of 0.1128 u, and converting this mass defect to energy by multiplying by 931 MeV/u, yielding 105.02 MeV.

Step-by-Step Solution

1

Determine the number of constituent protons and neutrons

Z = 7

protons and

N = 14 - 7 = 7

neutrons

The atomic number is 7 and the mass number is 14.

2

Calculate the total mass of the constituent nucleons

M_{\text{nucleons}} = (7 \times 1.0073\text{ u}) + (7 \times 1.0087\text{ u}) = 7.0511\text{ u} + 7.0609\text{ u} = 14.1120\text{ u}

The sum of the individual rest masses of all isolated protons and neutrons.

3

Compute the mass defect

\Delta m = 14.1120\text{ u} - 13.9992\text{ u} = 0.1128\text{ u}

Mass defect is the difference between total mass of isolated nucleons and the bound nuclear mass.

4

Convert the mass defect into energy in MeV

E_b = 0.1128\text{ u} \times 931\text{ MeV/u} = 105.0168\text{ MeV} \approx 105.02\text{ MeV}

Applying the mass-energy equivalence factor

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

.

Key Concept

Mass Defect and Binding Energy

Estimated Time:2m 0s

Question 104Question →

In a nuclear fusion reaction, a deuterium nucleus (

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

) fuses with a tritium nucleus (

{^{3}_{1}\text{H}}

) according to the equation:

{^{2}_{1}\text{H}} + {^{3}_{1}\text{H}} \rightarrow {^{4}_{2}\text{He}} + {^{1}_{0}\text{n}} + Q

Given the atomic masses:
- Mass of ${^{2}_{1}\text{H}} = 2.0141\text{ u}$
- Mass of ${^{3}_{1}\text{H}} = 3.0160\text{ u}$
- Mass of ${^{4}_{2}\text{He}} = 4.0026\text{ u}$
- Mass of ${^{1}_{0}\text{n}} = 1.0087\text{ u}$
- $1\text{ u} = 931.5\text{ MeV}$

What is the total energy $Q$ released in this reaction?

17.51\text{ MeV}

957.12\text{ MeV}

1.75\text{ MeV}

4686.38\text{ MeV}

Show answer & explanation

Answer: $17.51\text{ MeV}$

Answer

The total energy released in the reaction is

17.51\text{ MeV}

.

The correct answer of

17.51\text{ MeV}

is obtained by subtracting the combined mass of the helium-4 nucleus and neutron (

5.0113\text{ u}

) from the combined mass of the deuterium and tritium nuclei (

5.0301\text{ u}

) to yield a mass defect of

0.0188\text{ u}

, which equates to

17.51\text{ MeV}

when multiplied by

931.5\text{ MeV/u}

.

Step-by-Step Solution

1

Calculate the total mass of the reactants before the fusion reaction.

\text{Mass of reactants} = 2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}

To find the mass defect, we must first sum the masses of the initial nuclei.

2

Calculate the total mass of the reaction products.

\text{Mass of products} = 4.0026\text{ u} + 1.0087\text{ u} = 5.0113\text{ u}

The products include both the helium-4 nucleus and the emitted neutron.

3

Determine the mass defect (

\Delta m

).

\Delta m = 5.0301\text{ u} - 5.0113\text{ u} = 0.0188\text{ u}

The mass defect represents the mass converted into energy during fusion.

4

Convert the mass defect into released energy

Q

in

\text{MeV}

.

Q = 0.0188\text{ u} \times 931.5\text{ MeV/u} = 17.5122\text{ MeV} \approx 17.51\text{ MeV}

Using Einstein's mass-energy equivalence with

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

.

Key Concept

Mass defect and energy release in nuclear fusion reactions

Estimated Time:1m 30s

Question 105Question →

An electron of mass $9.0 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$ is accelerated from rest through an electric potential difference $V$ . If the associated de Broglie wavelength of the electron is $1.1 \times 10^{-10}\text{ m}$ and Planck's constant is $6.6 \times 10^{-34}\text{ J s}$ , what is the value of the potential difference $V$ ?

125\text{ V}

250\text{ V}

11.2\text{ V}

62.5\text{ V}

Show answer & explanation

Answer: $125\text{ V}$

Answer

The accelerating potential difference required is

125\text{ V}

.

The de Broglie wavelength of an electron accelerated from rest through a potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2meV}}

. Rearranging for

V

gives

V = \frac{h^2}{2me\lambda^2}

. Substituting

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

,

m = 9.0 \times 10^{-31}\text{ kg}

,

e = 1.6 \times 10^{-19}\text{ C}

, and

\lambda = 1.1 \times 10^{-10}\text{ m}

gives

V = 125\text{ V}

.

Step-by-Step Solution

1

Relate kinetic energy to momentum and accelerating potential.

E_k = eV

and

p = \sqrt{2m E_k} = \sqrt{2meV}

An electron accelerated through potential

V

gains kinetic energy equal to

eV

.

2

Substitute momentum into the de Broglie wavelength equation.

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}

De Broglie wavelength is defined as Planck's constant divided by momentum.

3

Rearrange the equation to solve for potential difference

V

.

V = \frac{h^2}{2me\lambda^2}

Squaring both sides allows isolating

V

.

4

Substitute the given numerical values and calculate

V

.

$V = \frac{(6.6 \times 10^{-34})^2}{2(9.0 \times 10^{-31})(1.6 \times 10^{-19})(1.1 \times 10^{-10})^2} = 125\text{ V}$

Performing the algebraic substitution yields the final potential.

Key Concept

de Broglie wavelength of an electron accelerated through a potential difference

Estimated Time:1m 30s

Question 106Question →

An $\alpha$ -particle has a mass approximately $4$ times that of a proton and carries $2$ times the elementary charge of a proton. If both particles are accelerated from rest through the same potential difference $V$ , what is the ratio of the de Broglie wavelength of the $\alpha$ -particle to that of the proton?

1 : 2\sqrt{2}

1 : 8

2\sqrt{2} : 1

1 : 2

Show answer & explanation

Answer: $1 : 2\sqrt{2}$

Answer

The ratio of the de Broglie wavelength of the

\alpha

-particle to that of the proton is

1 : 2\sqrt{2}

.

The de Broglie wavelength of a particle accelerated by a potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2mqV}}

. Taking the ratio of wavelengths yields

\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} = \sqrt{\frac{m_p e}{(4m_p)(2e)}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}

.

Step-by-Step Solution

1

Relate de Broglie wavelength to accelerating potential

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} = \frac{h}{\sqrt{2mqV}}

The kinetic energy gained by a charged particle accelerated through a potential difference

V

is

E_k = qV

.

2

Express wavelengths for both particles

\lambda_p = \frac{h}{\sqrt{2 m_p e V}}

and

\lambda_\alpha = \frac{h}{\sqrt{2 m_\alpha q_\alpha V}}

Setting up the parametric expressions for proton (

m_p, e

) and

\alpha

-particle (

m_\alpha, q_\alpha

).

3

Substitute relative values

m_\alpha = 4m_p

and

q_\alpha = 2e

\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{h}{\sqrt{16 m_p e V}} = \frac{h}{4\sqrt{m_p e V}}

Using the given relationships between mass and charge of the two particles.

4

Calculate the ratio

\frac{\lambda_\alpha}{\lambda_p}

\frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{4\sqrt{m_p e V}}}{\frac{h}{\sqrt{2 m_p e V}}} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}

Simplifying the fractional ratio of wavelengths.

Key Concept

de Broglie Wavelength of Charged Particles Accelerated in Electric Fields

Question 107Question →

A sample of a radioactive isotope used in medical imaging has an initial activity of $320\text{ MBq}$ . If its activity decreases to $20\text{ MBq}$ after an elapsed time of $15\text{ hours}$ , what is the half-life of the isotope in hours?

Show answer & explanation

Answer: 3.75

Answer

The half-life of the radioactive isotope is

3.75\text{ hours}

.

The initial activity of

320\text{ MBq}

drops to

20\text{ MBq}

, which is a reduction to

\frac{20}{320} = \frac{1}{16}

of its original value. Since

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

, exactly 4 half-lives have elapsed over the period of

15\text{ hours}

. Therefore, one half-life is

\frac{15\text{ hours}}{4} = 3.75\text{ hours}

.

Step-by-Step Solution

1

Calculate the fraction of initial activity remaining

Fraction remaining =

\frac{20\text{ MBq}}{320\text{ MBq}} = \frac{1}{16}

The ratio of current activity to initial activity determines the fraction of un-decayed nuclei remaining.

2

Determine the number of half-lives (

n

) that have elapsed

n = 4

half-lives, since

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

Each half-life reduces the remaining sample activity by half.

3

Calculate the half-life duration (

T_{1/2}

)

T_{1/2} = \frac{t}{n} = \frac{15\text{ hours}}{4} = 3.75\text{ hours}

Dividing total elapsed time by the number of half-lives gives the duration of one half-life.

Key Concept

Radioactive Decay Law and relationship between remaining activity fraction, number of half-lives, and total elapsed time.

Question 108Question →

In a nuclear fusion reaction, a lithium-6 nucleus ( ${^{6}_{3}\text{Li}}$ ) fuses with a deuterium nucleus ( ${^{2}_{1}\text{H}}$ ) to form two alpha particles ( ${^{4}_{2}\text{He}}$ ). Given the atomic masses of ${^{6}_{3}\text{Li}} = 6.0151\text{ u}$ , ${^{2}_{1}\text{H}} = 2.0141\text{ u}$ , and ${^{4}_{2}\text{He}} = 4.0026\text{ u}$ , and taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 22.356

Answer

The total energy released in the fusion reaction is 22.356 MeV.

The total mass before fusion (

{^{6}_{3}\text{Li}} + {^{2}_{1}\text{H}}

) is

8.0292\text{ u}

, while the total mass after fusion (

2 \times {^{4}_{2}\text{He}}

) is

8.0052\text{ u}

. The resulting mass defect is

\Delta m = 0.0240\text{ u}

. Multiplying this mass loss by

931.5\text{ MeV/u}

gives an energy release of

22.356\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total initial mass of the reactants

m_{\text{initial}} = 6.0151\text{ u} + 2.0141\text{ u} = 8.0292\text{ u}

Add the rest masses of the lithium-6 nucleus and the deuterium nucleus together.

2

Calculate the total final mass of the products

m_{\text{final}} = 2 \times 4.0026\text{ u} = 8.0052\text{ u}

The reaction produces two helium-4 nuclei (alpha particles), so multiply the mass of one alpha particle by 2.

3

Determine the mass defect

\Delta m = m_{\text{initial}} - m_{\text{final}} = 8.0292\text{ u} - 8.0052\text{ u} = 0.0240\text{ u}

Subtract the final mass of products from the initial mass of reactants to find the mass lost.

4

Convert the mass defect into energy released

E = \Delta m \times 931.5\text{ MeV/u} = 0.0240 \times 931.5 = 22.356\text{ MeV}

Apply Einstein's mass-energy equivalence principle using the standard conversion factor

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

.

Key Concept

Mass Defect and Energy Release in Nuclear Fusion Reactions

Question 109Question →

An $\alpha$ -particle (charge $+2e$ ) and a $\beta^-$ -particle (charge $-e$ ) emitted from a natural radioactive source enter a region of uniform electric field $E$ perpendicularly with equal initial kinetic energies. If $y_\alpha$ and $y_\beta$ represent the magnitudes of their transverse deflections after traveling the same horizontal distance through the field, what is the ratio $\frac{y_\beta}{y_\alpha}$ ?

\frac{1}{2}

2

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

1

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

The ratio of the transverse deflections

\frac{y_\beta}{y_\alpha}

is

\frac{1}{2}

.

For a charged particle entering a uniform electric field perpendicularly with kinetic energy

K

, the transverse deflection is

y = \frac{|q| E L^2}{4K}

. Because both particles enter with identical kinetic energies and travel the same horizontal distance

L

, the deflection is directly proportional to the magnitude of charge

|q|

and independent of mass. Since the

\beta^-

-particle has charge magnitude

e

and the

\alpha

-particle has charge magnitude

2e

, the ratio of their deflections is

\frac{e}{2e} = \frac{1}{2}

.

Step-by-Step Solution

1

Express initial horizontal velocity in terms of kinetic energy

K

and mass

m

.

v_x = \sqrt{\frac{2K}{m}}

, so time spent in the electric field of horizontal length

L

is

t = \frac{L}{v_x} = L \sqrt{\frac{m}{2K}}

.

Both particles enter horizontally with equal initial kinetic energy

K

.

2

Derive the formula for transverse deflection

y

in a uniform electric field

E

.

The transverse force is

F_y = |q|E

, giving acceleration

a_y = \frac{|q|E}{m}

. The deflection is

y = \frac{1}{2} a_y t^2 = \frac{1}{2} \left(\frac{|q|E}{m}\right) \left(\frac{L^2 m}{2K}\right) = \frac{|q|E L^2}{4K}

.

The particle mass

m

cancels out when time is expressed in terms of kinetic energy.

3

Calculate

y_\alpha

and

y_\beta

using their respective charge magnitudes

|q_\alpha| = 2e

and

|q_\beta| = e

.

y_\alpha = \frac{2e E L^2}{4K} = \frac{e E L^2}{2K}

and

y_\beta = \frac{e E L^2}{4K}

.

An

\alpha

-particle carries a charge of

+2e

while a

\beta^-

-particle carries a charge of

-e

.

4

Compute the ratio

\frac{y_\beta}{y_\alpha}

.

\frac{y_\beta}{y_\alpha} = \frac{\frac{e E L^2}{4K}}{\frac{e E L^2}{2K}} = \frac{1}{2}

.

Dividing

y_\beta

by

y_\alpha

cancels all terms except the ratio of charge magnitudes.

Key Concept

Deflection of charged radioactive emissions in a uniform electric field under equal kinetic energy

Question 110Question →

The threshold wavelength for photoelectric emission from a metallic emitter is $\lambda_0$ . When light of wavelength $\frac{\lambda_0}{3}$ illuminates the emitter, photoelectrons are ejected with a maximum kinetic energy of $K_1$ . If the incident light is changed to a wavelength of $\frac{\lambda_0}{4}$ , what is the new maximum kinetic energy $K_2$ of the ejected photoelectrons in terms of $K_1$ ?

\frac{3}{2} K_1

\frac{4}{3} K_1

2 K_1

\frac{2}{3} K_1

Show answer & explanation

Answer: $\frac{3}{2} K_1$

Answer

The new maximum kinetic energy

K_2

is equal to

\frac{3}{2} K_1

.

According to Einstein's photoelectric equation, maximum kinetic energy is

K = E - W_0

. With threshold wavelength

\lambda_0

, the work function is

W_0 = \frac{hc}{\lambda_0}

. For light of wavelength

\frac{\lambda_0}{3}

, the photon energy is

3W_0

, yielding

K_1 = 3W_0 - W_0 = 2W_0

. For light of wavelength

\frac{\lambda_0}{4}

, the photon energy is

4W_0

, yielding

K_2 = 4W_0 - W_0 = 3W_0

. Comparing

K_2

and

K_1

gives

K_2 = \frac{3}{2} K_1

.

Step-by-Step Solution

1

Express the work function

W_0

in terms of threshold wavelength

\lambda_0

.

W_0 = \frac{hc}{\lambda_0}

The threshold wavelength defines the minimum photon energy required to liberate an electron from the metal surface.

2

Apply Einstein's photoelectric equation for the first incident wavelength

\lambda_1 = \frac{\lambda_0}{3}

.

K_1 = \frac{hc}{\lambda_0/3} - W_0 = 3\left(\frac{hc}{\lambda_0}\right) - W_0 = 3W_0 - W_0 = 2W_0

Einstein's photoelectric equation states that maximum kinetic energy equals photon energy minus work function.

3

Apply Einstein's photoelectric equation for the second incident wavelength

\lambda_2 = \frac{\lambda_0}{4}

.

K_2 = \frac{hc}{\lambda_0/4} - W_0 = 4\left(\frac{hc}{\lambda_0}\right) - W_0 = 4W_0 - W_0 = 3W_0

Calculate maximum kinetic energy for the shorter incident wavelength.

4

Determine the ratio of

K_2

to

K_1

.

\frac{K_2}{K_1} = \frac{3W_0}{2W_0} = \frac{3}{2} \implies K_2 = \frac{3}{2} K_1

Express the new kinetic energy in terms of the initial kinetic energy.

Key Concept

Einstein's Photoelectric Equation and Work Function Threshold

Question 111Question →

According to de Broglie's hypothesis on wave-particle duality, how does the de Broglie wavelength of a moving particle change if its momentum is doubled?

It is doubled

It is halved

It is quadrupled

It remains unchanged

Show answer & explanation

Answer: It is halved

Answer

The de Broglie wavelength is halved when the momentum of the particle is doubled.

By de Broglie's equation

\lambda = \frac{h}{p}

, the wavelength

\lambda

is inversely proportional to momentum

p

. Doubling the momentum replaces

p

with

2p

, resulting in a new wavelength of

\frac{h}{2p} = \frac{\lambda}{2}

, which means the wavelength is halved.

Step-by-Step Solution

1

State the de Broglie wavelength formula

\lambda = \frac{h}{p}

The de Broglie wavelength

\lambda

is inversely proportional to particle momentum

p

.

2

Substitute doubled momentum

p' = 2p

into the formula

\lambda' = \frac{h}{2p} = \frac{1}{2}\lambda

Increasing momentum by a factor of 2 scales the wavelength by

\frac{1}{2}

.

Key Concept

Inverse proportionality between de Broglie wavelength and particle momentum

Estimated Time:45s

Question 112Question →

An oxygen nucleus $^{16}_{8}\text{O}$ has a measured nuclear mass of $15.9906\text{ u}$ . Given that the mass of a proton is $1.0078\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , calculate the total binding energy of the nucleus in $\text{MeV}$ . (Take $1\text{ u} = 931.5\text{ MeV}$ )

Show answer & explanation

Answer: 131.71

Answer

The total binding energy of the oxygen nucleus is

131.71\text{ MeV}

.

Summing the masses of 8 individual protons and 8 individual neutrons gives

16.1320\text{ u}

. Subtracting the actual nuclear mass of

15.9906\text{ u}

leaves a mass defect of

0.1414\text{ u}

. Multiplying this mass defect by the equivalence constant

931.5\text{ MeV/u}

yields a total binding energy of

131.71\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total combined mass of the free constituent nucleons

Mass of 8 protons + 8 neutrons =

8(1.0078\text{ u}) + 8(1.0087\text{ u}) = 8.0624\text{ u} + 8.0696\text{ u} = 16.1320\text{ u}

Oxygen-16 has

Z = 8

protons and

A - Z = 16 - 8 = 8

neutrons.

2

Calculate the mass defect (

\Delta m

)

\Delta m = 16.1320\text{ u} - 15.9906\text{ u} = 0.1414\text{ u}

Mass defect is the difference between total constituent mass and actual nuclear mass.

3

Convert mass defect into total binding energy (

E_b

)

E_b = 0.1414\text{ u} \times 931.5\text{ MeV/u} = 131.7141\text{ MeV} \approx 131.71\text{ MeV}

Applying the energy equivalent factor of

931.5\text{ MeV}

per atomic mass unit.

Key Concept

Mass defect and nuclear binding energy equivalence

Question 113Question →

Which of the following natural radioactive emissions has the least penetrating power and can be stopped by a thin sheet of paper?

Alpha particles

Beta particles

Gamma rays

Neutrons

Show answer & explanation

Answer: Alpha particles

Answer

Alpha particles have the lowest penetrating power of the three primary natural radioactive emissions and are stopped by a thin sheet of paper.

Alpha particles consist of heavy, doubly charged helium nuclei (

^4_2\text{He}

). Their relatively large size and strong positive charge lead to frequent ionizing collisions with surrounding atoms, causing them to deposit their energy over a very short distance and making them unable to pass through a simple sheet of paper.

Step-by-Step Solution

1

Identify the nature and physical properties of natural radioactive emissions (alpha particles, beta particles, and gamma rays).

Alpha particles consist of 2 protons and 2 neutrons (helium nucleus,

^4_2\text{He}

), carrying a heavy mass (

4\text{ u}

) and a

+2e

charge.

Large mass and high charge increase the likelihood of collisions with atoms in matter.

2

Compare ionizing ability and penetrating power across emission types.

Because alpha particles cause high ionization over short distances, they lose kinetic energy rapidly.

High rate of energy loss corresponds directly to very low penetrating depth.

3

Determine the stopping material required for alpha particles.

Alpha particles are absorbed by a thin layer of matter, such as a single sheet of paper or a few centimeters of air.

This confirms alpha radiation has the least penetrating power among natural emissions.

Key Concept

Relative Penetrating and Ionizing Capacities of Natural Radioactive Emissions

Estimated Time:45s

Question 114Question →

An $\alpha$ -particle and a $\beta^-$ -particle emitted from a radioactive source enter a uniform magnetic field perpendicularly with equal linear momenta. What is the ratio of the radius of curvature of the trajectory of the $\alpha$ -particle to that of the $\beta^-$ -particle in the magnetic field?

1 : 2

2 : 1

4 : 1

1 : 4

Show answer & explanation

Answer: $1 : 2$

Answer

The ratio of the radius of curvature of the path of the

\alpha

-particle to that of the

\beta^-

-particle is

1 : 2

.

In a uniform magnetic field, the radius of curvature

r

of a moving charged particle is given by

r = \frac{p}{qB}

, where

p

is momentum,

q

is charge magnitude, and

B

is magnetic field flux density. Since both particles have equal momentum in the same field,

r \propto \frac{1}{q}

. The

\alpha

-particle carries charge magnitude

2e

while the

\beta^-

-particle carries charge magnitude

e

. Consequently, the ratio of their radii is

r_\alpha : r_\beta = \frac{1}{2} : 1 = 1 : 2

.

Step-by-Step Solution

1

Relate radius of trajectory to linear momentum and magnetic field

The magnetic force supplies centripetal force:

qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} = \frac{p}{qB}

, where

p

is linear momentum.

Expressing radius in terms of momentum directly utilizes the given condition that

p

is equal for both particles.

2

Identify the magnitude of charge for each emission

For the

\alpha

-particle (

^4_2\text{He}^{2+}

),

q_\alpha = 2e

. For the

\beta^-

-particle (

^0_{-1}\text{e}

),

q_\beta = e

.

Radius of curvature depends on the magnitude of charge carried by each radiation type.

3

Calculate the ratio of radii

r_\alpha / r_\beta

\frac{r_\alpha}{r_\beta} = \frac{\frac{p}{2eB}}{\frac{p}{eB}} = \frac{e}{2e} = \frac{1}{2}.

Since momentum

p

and field strength

B

are identical, the ratio simplifies directly to the inverse ratio of their charge magnitudes.

Key Concept

Deflection of radioactive emissions in magnetic fields and radius of curvature

Question 115Question →

A radioactive sample has a half-life of $5\text{ days}$ . If the mass of the sample that has decayed after $20\text{ days}$ is $45\text{ g}$ , what was the initial mass of the sample?

48\text{ g}

720\text{ g}

60\text{ g}

180\text{ g}

Show answer & explanation

Answer: $48\text{ g}$

Answer

The initial mass of the sample was

48\text{ g}

.

The elapsed time corresponds to 4 half-lives (

20 / 5 = 4

). The remaining mass fraction is

(1/2)^4 = 1/16

, which means

15/16

of the original mass has decayed. Setting

15/16

of the initial mass equal to

45\text{ g}

gives an initial mass of

48\text{ g}

.

Step-by-Step Solution

1

Calculate the number of half-lives elapsed

n = \frac{t}{T_{1/2}} = \frac{20\text{ days}}{5\text{ days}} = 4\text{ half-lives}

Determining how many half-life cycles occurred within the total elapsed time.

2

Determine the fraction of sample decayed

Fraction remaining =

(1/2)^4 = 1/16

; Fraction decayed =

1 - 1/16 = 15/16

The decayed fraction is the complement of the remaining fraction.

3

Calculate the initial mass

M_{\text{decayed}} = \frac{15}{16} M_0 = 45\text{ g} \implies M_0 = 45 \times \frac{16}{15} = 48\text{ g}

Relating the given decayed mass to the total initial mass.

Key Concept

Radioactive Decay Law and Half-life

Question 116Question →

In a nuclear fission reaction, a Uranium-235 nucleus absorbs a neutron and splits into Tellurium-137 and Zirconium-97, releasing energy. If the total mass of the reactants is $236.053\text{ u}$ and the total mass of the products is $235.853\text{ u}$ , what is the energy released in $\text{MeV}$ ? (Take $1\text{ u} = 931\text{ MeV}$ )

18.62\text{ MeV}

186.20\text{ MeV}

0.000215\text{ MeV}

186.20\text{ J}

Show answer & explanation

Answer: $186.20\text{ MeV}$

Answer

The total energy released in the fission reaction is

186.20\text{ MeV}

.

The energy released during nuclear fission is calculated from the mass defect (

\Delta m

). Subtracting product mass from reactant mass gives

\Delta m = 236.053\text{ u} - 235.853\text{ u} = 0.200\text{ u}

. Multiplying this mass defect by the standard conversion factor (

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

) yields

0.200 \times 931 = 186.20\text{ MeV}

.

Step-by-Step Solution

1

Calculate the mass defect (

\Delta m

)

\Delta m = 236.053\text{ u} - 235.853\text{ u} = 0.200\text{ u}

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

2

Convert the mass defect into energy in

\text{MeV}

E = 0.200\text{ u} \times 931\text{ MeV/u} = 186.20\text{ MeV}

According to mass-energy equivalence, each atomic mass unit (

\text{u}

) liberates

931\text{ MeV}

of energy.

Key Concept

Mass Defect and Energy Release in Nuclear Fission

Question 117Question →

Light of frequency $8.0 \times 10^{14}\text{ Hz}$ illuminates a photosensitive plate, causing photoelectrons to be emitted with a maximum kinetic energy of $1.2\text{ eV}$ . If the same plate is subsequently illuminated by light of frequency $1.2 \times 10^{15}\text{ Hz}$ , what is the stopping potential, in volts, needed to reduce the photoelectric current to zero? (Take $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 2.85

Answer

The stopping potential needed to reduce the photoelectric current to zero is 2.85 V.

Using Einstein's photoelectric equation

E = W_0 + K_{\text{max}}

, the initial photon energy is

E_1 = h f_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

. Given

K_1 = 1.2\text{ eV}

, the work function of the metal is

W_0 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}

. For the second frequency

f_2 = 1.2 \times 10^{15}\text{ Hz}

, the photon energy is

E_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}

. The new maximum kinetic energy is

K_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}

. Since

e V_s = K_{\text{max}}

, the stopping potential required to reduce the current to zero is

2.85\text{ V}

.

Step-by-Step Solution

1

Calculate the photon energy

E_1

of the initial light in electron-volts

E_1 = \frac{6.6 \times 10^{-34} \times 8.0 \times 10^{14}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

Photon energy is related to frequency by

E = h f

.

2

Determine the work function

W_0

of the photosensitive plate

W_0 = E_1 - K_1 = 3.3\text{ eV} - 1.2\text{ eV} = 2.1\text{ eV}

By Einstein's photoelectric equation,

K_{\text{max}} = E - W_0

.

3

Calculate the photon energy

E_2

for the second light frequency

E_2 = \frac{6.6 \times 10^{-34} \times 1.2 \times 10^{15}}{1.6 \times 10^{-19}} = 4.95\text{ eV}

The energy of the second photon is calculated using

f_2 = 1.2 \times 10^{15}\text{ Hz}

.

4

Find the maximum kinetic energy

K_2

and corresponding stopping potential

V_s

K_2 = 4.95\text{ eV} - 2.1\text{ eV} = 2.85\text{ eV}

, giving

V_s = 2.85\text{ V}

The stopping potential in volts is numerical equal to the maximum kinetic energy expressed in electron-volts (

e V_s = K_{\text{max}}

).

Key Concept

Einstein's Photoelectric Equation and Stopping Potential

Estimated Time:2m 0s

Question 118Question →

A deuteron of mass $2m_p$ and charge $e$ , and an $\alpha$ -particle of mass $4m_p$ and charge $2e$ , are accelerated from rest through potential differences of $V_d$ and $V_\alpha$ respectively. If both particles acquire equal de Broglie wavelengths, what is the ratio of their accelerating potential differences, $\frac{V_d}{V_\alpha}$ ?

4 : 1

1 : 4

2 : 1

16 : 1

Show answer & explanation

Answer: $4 : 1$

Answer

The ratio of the accelerating potential differences

\frac{V_d}{V_\alpha}

is

4 : 1

.

Using the de Broglie wavelength equation

\lambda = \frac{h}{\sqrt{2mqV}}

, setting

\lambda_d = \lambda_\alpha

gives

m_d q_d V_d = m_\alpha q_\alpha V_\alpha

. Substituting

m_d = 2m_p

,

q_d = e

,

m_\alpha = 4m_p

, and

q_a = 2e

yields

2 V_d = 8 V_\alpha

, which simplifies directly to

\frac{V_d}{V_\alpha} = 4 : 1

.

Step-by-Step Solution

1

Express the de Broglie wavelength in terms of particle mass, charge, and accelerating potential difference.

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}

Kinetic energy gained by a charged particle in an electric field is

E_k = qV

, and momentum is

p = \sqrt{2mE_k} = \sqrt{2mqV}

.

2

Set the de Broglie wavelengths of the deuteron and

\alpha

-particle equal to each other.

\frac{h}{\sqrt{2 m_d q_d V_d}} = \frac{h}{\sqrt{2 m_\alpha q_\alpha V_\alpha}} \implies m_d q_d V_d = m_\alpha q_\alpha V_\alpha

Equal wavelengths mean their denominators in the de Broglie expression must be equal.

3

Substitute the given values for mass and charge into the equality.

(2m_p)(e) V_d = (4m_p)(2e) V_\alpha \implies 2 m_p e V_d = 8 m_p e V_\alpha

Deuteron has mass

2m_p

and charge

e

;

\alpha

-particle has mass

4m_p

and charge

2e

.

4

Solve for the ratio

\frac{V_d}{V_\alpha}

.

\frac{V_d}{V_\alpha} = \frac{8}{2} = 4

Simplifying the algebraic equation by cancelling common terms

m_p

and

e

.

Key Concept

De Broglie Wavelength of Accelerated Charged Particles

Question 119Question →

A subatomic particle of mass $2.0 \times 10^{-27}\text{ kg}$ possesses a kinetic energy of $1.0 \times 10^{-19}\text{ J}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , what is the de Broglie wavelength of the particle?

3.3 \times 10^{-11}\text{ m}

4.7 \times 10^{-11}\text{ m}

1.65 \times 10^{-11}\text{ m}

3.3 \times 10^{-57}\text{ m}

Show answer & explanation

Answer: $3.3 \times 10^{-11}\text{ m}$

Answer

3.3 \times 10^{-11}\text{ m}

The momentum of the particle is given by

p = \sqrt{2mE_k} = \sqrt{2 \times 2.0 \times 10^{-27} \times 1.0 \times 10^{-19}} = 2.0 \times 10^{-23}\text{ kg m/s}

. Substituting into the de Broglie wavelength relation

\lambda = \frac{h}{p}

gives

\lambda = \frac{6.6 \times 10^{-34}}{2.0 \times 10^{-23}} = 3.3 \times 10^{-11}\text{ m}

.

Step-by-Step Solution

1

Relate kinetic energy to momentum

p = \sqrt{2mE_k} = \sqrt{2 \times (2.0 \times 10^{-27}\text{ kg}) \times (1.0 \times 10^{-19}\text{ J})} = \sqrt{4.0 \times 10^{-46}} = 2.0 \times 10^{-23}\text{ kg m/s}

The de Broglie wavelength formula requires linear momentum, which connects to kinetic energy via

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

.

2

Calculate the de Broglie wavelength

\lambda = \frac{h}{p} = \frac{6.6 \times 10^{-34}\text{ J s}}{2.0 \times 10^{-23}\text{ kg m/s}} = 3.3 \times 10^{-11}\text{ m}

Applying de Broglie's wave-particle duality relation

\lambda = \frac{h}{p}

.

Key Concept

de Broglie Wavelength and Kinetic Energy Relation

Estimated Time:1m 30s

Question 120Question →

A lithium-6 nucleus $^{6}_{3}\text{Li}$ has a measured nuclear mass of $6.0151\text{ u}$ . Given that the mass of a proton is $1.0078\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , calculate the binding energy per nucleon of the nucleus in $\text{MeV}$ . (Take $1\text{ u} = 931\text{ MeV}$ )

Show answer & explanation

Answer: 5.34

Answer

The binding energy per nucleon of the

^{6}_{3}\text{Li}

nucleus is approximately

5.34\text{ MeV}

(or

5.34\text{ MeV/nucleon}

).

The total mass of 3 free protons and 3 free neutrons is

6.0495\text{ u}

. Subtracting the actual nuclear mass (

6.0151\text{ u}

) yields a mass defect of

0.0344\text{ u}

. Multiplying by

931\text{ MeV/u}

gives a total binding energy of

32.0264\text{ MeV}

. Dividing by the

6

nucleons in lithium-6 yields

5.34\text{ MeV}

per nucleon.

Step-by-Step Solution

1

Determine the number of protons and neutrons and compute the total constituent mass.

Protons

Z = 3

, Neutrons

N = 3

. Total nucleon mass

m_{\text{nucleons}} = 3(1.0078\text{ u}) + 3(1.0087\text{ u}) = 6.0495\text{ u}

.

Free nucleons have a combined mass greater than the bound nucleus.

2

Calculate the mass defect (

\Delta m

).

\Delta m = 6.0495\text{ u} - 6.0151\text{ u} = 0.0344\text{ u}

.

The difference between total constituent mass and measured nuclear mass represents the lost mass converted into binding energy.

3

Convert mass defect to total binding energy in

\text{MeV}

.

E_b = 0.0344\text{ u} \times 931\text{ MeV/u} = 32.0264\text{ MeV}

.

Using the equivalence

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

.

4

Calculate binding energy per nucleon by dividing by mass number

A = 6

.

\frac{32.0264\text{ MeV}}{6} = 5.3377\text{ MeV} \approx 5.34\text{ MeV}.

Binding energy per nucleon measures the stability per particle in the nucleus.

Key Concept

Mass Defect and Binding Energy per Nucleon

Atomic and Nuclear Physics

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