Atomic and Nuclear Physics Alıştırma Soruları — JAMB UTME — Sayfa 3

Soru 41Soru →

In J. J. Thomson's plum pudding model of the atom, how is the positive charge distributed throughout the atom?

It is uniformly spread throughout a spherical volume containing embedded electrons.

It is concentrated in a tiny, dense central core called the nucleus.

It orbits around a central cluster of negative charges in fixed energy levels.

It is localized entirely on the outer surface of a hollow atomic shell.

Cevabı ve açıklamayı göster

Cevap: It is uniformly spread throughout a spherical volume containing embedded electrons.

Cevap

The positive charge is uniformly spread throughout a spherical volume containing embedded electrons.

In J. J. Thomson's plum pudding model, the atom is envisioned as a continuous sphere of positive charge within which negatively charged electrons are embedded evenly to maintain overall electrical neutrality.

Adım Adım Çözüm

1

Recall the fundamental proposal of Thomson's atomic model (1897).

J. J. Thomson envisioned the atom as a sphere of positive electrification.

This accounted for atomic neutrality following his discovery of the electron.

2

Identify the arrangement of electrons within this positive sphere.

Electrons were thought to be embedded throughout the positive sphere like plums in a pudding.

This balanced the positive charge uniformly across the atomic volume.

Anahtar Kavram

Thomson's Plum Pudding Atomic Model

Tahmini Süre:45s

Soru 42Soru →

In Bohr's model of the hydrogen atom, the radius of a stationary orbit is proportional to $n^2$ , where $n$ is the principal quantum number. If the radius of the ground-state orbit ( $n = 1$ ) is $r_1$ , what is the radius of the orbit corresponding to the second excited state?

3r_1

4r_1

6r_1

9r_1

Cevabı ve açıklamayı göster

Cevap: $9r_1$

Cevap

The radius of the electron's orbit in the second excited state is

9r_1

.

In Bohr's atomic model, the ground state corresponds to the quantum number

n = 1

. The excited states are numbered sequentially above the ground state: the first excited state is

n = 2

and the second excited state is

n = 3

. Since the orbital radius scales with the square of the principal quantum number (

r_n = n^2 r_1

), substituting

n = 3

yields

r_3 = 3^2 r_1 = 9r_1

. Thus, the option stating

9r_1

is correct.

Adım Adım Çözüm

1

Identify the principal quantum number

n

for the second excited state.

n = 3

The ground state corresponds to

n = 1

, the first excited state corresponds to

n = 2

, and the second excited state corresponds to

n = 3

.

2

Apply Bohr's radius formula for hydrogen-like atoms.

r_n = n^2 r_1

According to Bohr's postulates, the orbital radius is proportional to the square of the principal quantum number

n

.

3

Calculate

r_3

by substituting

n = 3

into the radius expression.

r_3 = 3^2 r_1 = 9r_1

Squaring

n = 3

gives

9

, making the radius 9 times the ground-state radius.

Anahtar Kavram

Bohr's quantization of orbital radius (

r_n \propto n^2

)

Tahmini Süre:1m 0s

Soru 43Soru →

An electron in a hydrogen atom transitions from an excited energy state of $-3.4\text{ eV}$ to the ground state of $-13.6\text{ eV}$ . What is the energy of the photon emitted during this transition?

10.2\text{ eV}

17.0\text{ eV}

-10.2\text{ eV}

3.4\text{ eV}

Cevabı ve açıklamayı göster

Cevap: $10.2\text{ eV}$

Cevap

10.2\text{ eV}

When an electron drops to a lower energy state, the energy of the emitted photon equals the difference between the initial and final energy levels:

E_{photon} = E_{initial} - E_{final} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV}

.

Adım Adım Çözüm

1

Identify the initial and final energy states.

Initial state

E_i = -3.4\text{ eV}

and final ground state

E_f = -13.6\text{ eV}

.

The electron drops from the higher energy state to the lower state.

2

Calculate the photon energy using the transition formula

E_{photon} = E_i - E_f

.

E_{photon} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV}

.

By energy conservation, the energy of the emitted photon must equal the energy loss of the electron.

Anahtar Kavram

Photon Emission in Atomic Energy Level Transitions

Tahmini Süre:45s

Soru 44Soru →

Historical developments in atomic physics led to several distinct models of atomic structure. Match each atomic model on the left with its defining structural feature or experimental basis on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Thomson's Model matches with embedded electrons in a positive sphere; Rutherford's Model matches with a dense positive nucleus from alpha scattering; Bohr's Model matches with quantized non-radiating orbits; Quantum Mechanical Model matches with electron probability orbitals.

Each historical atomic model is correctly matched to its fundamental feature: Thomson proposed electrons suspended in positive mass; Rutherford used alpha scattering to discover the compact nucleus; Bohr quantized electron orbits to explain spectral lines; and the Quantum Mechanical Model represents electrons via three-dimensional probability orbitals.

Adım Adım Çözüm

1

Identify Thomson's contribution

Thomson proposed the 'plum pudding' model where negative electrons are embedded inside a uniform positive charge sphere.

This preceded the discovery of the atomic nucleus.

2

Identify Rutherford's contribution

Rutherford discovered the central positive nucleus through the alpha particle deflection experiment.

Large deflections meant most atomic mass and positive charge concentrated at a tiny core.

3

Identify Bohr's contribution

Bohr added quantum conditions to planetary orbits so electrons remain stable without continuously radiating energy.

Quantized angular momentum explains discrete emission line spectra.

4

Identify the Quantum Mechanical Model contribution

Modern quantum mechanics replaces fixed circular orbits with wave functions and 3D probability clouds (orbitals).

Heisenberg's uncertainty principle rules out precise circular orbits.

Anahtar Kavram

Evolution of Atomic Models

Soru 45Soru →

An electron in an excited atom transitions from an energy level of $-2.5\text{ eV}$ to a lower energy level of $-8.5\text{ eV}$ . What is the energy of the emitted photon in Joules? ( $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

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

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

1.76 \times 10^{-18}\text{ J}

-9.6 \times 10^{-19}\text{ J}

Cevabı ve açıklamayı göster

Cevap: $9.6 \times 10^{-19}\text{ J}$

Cevap

The energy of the emitted photon is

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

.

The energy of the emitted photon is given by

\Delta E = E_i - E_f = -2.5\text{ eV} - (-8.5\text{ eV}) = 6.0\text{ eV}

. Converting this to Joules gives

6.0 \times 1.6 \times 10^{-19}\text{ J} = 9.6 \times 10^{-19}\text{ J}

.

Adım Adım Çözüm

1

Calculate the energy difference

\Delta E

between the initial and final energy levels.

\Delta E = E_{\text{initial}} - E_{\text{final}} = -2.5\text{ eV} - (-8.5\text{ eV}) = 6.0\text{ eV}

.

When an electron drops to a lower energy state, it emits a photon with energy equal to the difference between the two energy levels.

2

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

E = 6.0\text{ eV} \times 1.6 \times 10^{-19}\text{ J/eV} = 9.6 \times 10^{-19}\text{ J}

.

Standard SI unit calculations require multiplying the value in eV by

1.6 \times 10^{-19}\text{ J/eV}

.

Anahtar Kavram

Photon Emission during Atomic Transitions

Tahmini Süre:45s

Soru 46Soru →

Which of the following observations occurs when a beam of cathode rays passes between two oppositely charged parallel metal plates?

The beam is deflected towards the positively charged plate.

The beam is deflected towards the negatively charged plate.

The beam passes straight through without any deflection.

The beam splits into two components moving toward both plates.

Cevabı ve açıklamayı göster

Cevap: The beam is deflected towards the positively charged plate.

Cevap

The beam is deflected towards the positively charged plate.

Cathode rays consist of negatively charged electrons. When placed in an electric field between two oppositely charged plates, the electrostatic attraction pulls the negatively charged electrons toward the positively charged plate.

Adım Adım Çözüm

1

Identify the nature and electric charge of cathode rays.

Cathode rays are streams of fast-moving electrons, which carry a negative electric charge.

Determining the sign of the charge is required to find the direction of the electrostatic force.

2

Apply electrostatic force principles to the beam in an electric field.

Opposite charges attract, so negatively charged electrons experience a force toward the positively charged plate.

An electric field exerts an attractive force on negative charges directed toward the positive region.

Anahtar Kavram

Deflection of Cathode Rays in an Electric Field

Soru 47Soru →

In a mercury vapor tube, an excited atom transitions from an upper energy level of $-3.71\text{ eV}$ to a lower energy level of $-5.54\text{ eV}$ . What is the frequency of the emitted photon? (Take $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $4.44 \times 10^{14}\text{ Hz}$

Cevap

The frequency of the emitted photon is

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

.

The energy lost by the atom during transition is

\Delta E = -3.71\text{ eV} - (-5.54\text{ eV}) = 1.83\text{ eV}

. Converting to Joules gives

1.83 \times 1.6 \times 10^{-19}\text{ J} = 2.928 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant (

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

) yields a photon frequency of

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

.

Adım Adım Çözüm

1

Calculate the energy difference (

\Delta E

) between the two atomic energy levels.

\Delta E = E_{\text{initial}} - E_{\text{final}} = -3.71\text{ eV} - (-5.54\text{ eV}) = 1.83\text{ eV}

The energy of the emitted photon equals the difference in energy between the initial and final states.

2

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

\Delta E = 1.83 \times 1.6 \times 10^{-19}\text{ J} = 2.928 \times 10^{-19}\text{ J}

Planck's constant is given in SI units (J s), so the energy must be in Joules.

3

Apply Planck's equation

E = hf

to find the photon frequency

f

.

f = \frac{\Delta E}{h} = \frac{2.928 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}} \approx 4.44 \times 10^{14}\text{ Hz}

The frequency of an emitted photon is directly proportional to its energy difference.

Anahtar Kavram

Atomic transition photon frequency calculation (

E = hf = E_2 - E_1

).

Tahmini Süre:1m 30s

Soru 48Soru →

According to Bohr's atomic model of the hydrogen atom, what condition must be satisfied by an electron moving in a stable stationary orbit?

Its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Its total energy is proportional to light intensity.

Its orbital frequency increases continuously as it radiates energy in the orbit.

Its total mechanical energy is zero in all allowed stationary orbits.

Cevabı ve açıklamayı göster

Cevap: Its orbital angular momentum is an integral multiple of $\frac{h}{2\pi}$ .

Cevap

An electron moves in a stable stationary orbit when its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Bohr's fundamental postulate states that an electron can revolve around the nucleus only in certain non-radiating orbits (stationary states) where its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Adım Adım Çözüm

1

Recall Bohr's postulates for the hydrogen atom

Identify that stable orbits require quantization of orbital angular momentum.

Bohr introduced quantization to explain why orbiting electrons do not continuously emit radiation and collapse into the nucleus.

2

State the mathematical formula for angular momentum quantization

L = mvr = \frac{nh}{2\pi}

, where

n

is an integer (

1, 2, 3, \dots

) and

h

is Planck's constant.

Only specific discreet orbits meeting this condition are allowed stationary states.

Anahtar Kavram

Quantization of Angular Momentum in Bohr's Model

Tahmini Süre:45s

Soru 49Soru →

In Bohr's atomic model of the hydrogen atom, the radius of the ground state orbit ( $n = 1$ ) is $0.053\text{ nm}$ . According to de Broglie's condition for stationary electron orbits, what is the de Broglie wavelength of the electron in its second excited state? Express your answer in nanometers ( $\text{nm}$ ).

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The de Broglie wavelength of the electron in its second excited state is

1.00\text{ nm}

.

The second excited state corresponds to the quantum number

n = 3

. According to Bohr's atomic model, the radius of the

n

-th orbit is given by

r_n = n^2 r_1

, yielding

r_3 = 3^2 \times 0.053\text{ nm} = 0.477\text{ nm}

. De Broglie explained Bohr's angular momentum quantization by showing that an integral number of electron matter-waves must fit around the orbital circumference:

2\pi r_n = n \lambda_n

. Solving for

\lambda_3

gives

\lambda_3 = \frac{2\pi r_3}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 1.00\text{ nm}

.

Adım Adım Çözüm

1

Identify the principal quantum number for the specified energy state

n = 3

The ground state corresponds to

n = 1

, the first excited state to

n = 2

, and the second excited state to

n = 3

.

2

Calculate the radius of the third stationary orbit

r_3 = 0.477\text{ nm}

In Bohr's model, the radius of the

n

-th orbit is proportional to

n^2

, so

r_3 = 3^2 \times 0.053\text{ nm} = 9 \times 0.053\text{ nm} = 0.477\text{ nm}

.

3

Apply de Broglie's standing wave condition for stationary orbits

\lambda_3 = \frac{2\pi r_3}{3}

De Broglie postulated that a stationary orbit contains an integral number of electron de Broglie wavelengths around its circumference:

2\pi r_n = n \lambda_n

.

4

Substitute values to compute the wavelength

\lambda_3 = 1.00\text{ nm}

\lambda_3 = \frac{2 \times 3.1416 \times 0.477\text{ nm}}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 0.9992\text{ nm} \approx 1.00\text{ nm}

.

Anahtar Kavram

De Broglie Standing Wave Quantization in Bohr Atomic Model

Soru 50Soru →

During a head-on collision in Rutherford's $\alpha$ -particle scattering experiment, an $\alpha$ -particle of initial speed $v$ approaches a stationary heavy nucleus. The distance of closest approach achieved by the $\alpha$ -particle is $r_0$ . If the initial speed of the $\alpha$ -particle is increased to $2v$ , what will be the new distance of closest approach in terms of $r_0$ ?

\frac{r_0}{4}

\frac{r_0}{2}

2r_0

4r_0

Cevabı ve açıklamayı göster

Cevap: $\frac{r_0}{4}$

Cevap

The new distance of closest approach will be

\frac{r_0}{4}

.

At the distance of closest approach, the entire initial kinetic energy of the

\alpha

-particle is converted into electric potential energy:

E_k = \frac{1}{2}mv^2 = \frac{k_e q_1 q_2}{r}

. Rearranging for

r

gives

r = \frac{2k_e q_1 q_2}{m v^2}

, showing that

r

is inversely proportional to

v^2

. When the initial speed is doubled to

2v

, the kinetic energy increases by a factor of

2^2 = 4

. Consequently, the distance of closest approach is reduced to one-fourth of its initial value,

\frac{r_0}{4}

.

Adım Adım Çözüm

1

Apply conservation of mechanical energy at the distance of closest approach.

Initial kinetic energy equals electrostatic potential energy at distance

r_0

:

\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_0}

.

At the distance of closest approach, the

\alpha

-particle momentarily stops, converting all kinetic energy into electric potential energy.

2

Express the distance of closest approach

r_0

in terms of initial speed

v

.

r_0 = \frac{2 q_1 q_2}{4\pi\varepsilon_0 m v^2} \propto \frac{1}{v^2}

.

Rearranging the energy conservation equation demonstrates that distance of closest approach is inversely proportional to

v^2

.

3

Substitute the new speed

v' = 2v

into the proportion.

r' \propto \frac{1}{(2v)^2} = \frac{1}{4v^2} = \frac{r_0}{4}

.

Doubling the speed quadruples the kinetic energy, reducing the distance required to bring the particle to rest by a factor of 4.

Anahtar Kavram

Distance of Closest Approach in Rutherford Scattering

Tahmini Süre:2m 0s

Soru 51Soru →

Which of the following observations demonstrates that cathode rays possess kinetic energy and momentum?

The rotation of a light paddle wheel placed directly in their path

The casting of a sharp shadow when an opaque Maltese cross is placed in their path

The production of fluorescence when the rays strike the glass walls of the discharge tube

The deflection of the rays toward a positively charged plate in an electric field

Cevabı ve açıklamayı göster

Cevap: The rotation of a light paddle wheel placed directly in their path

Cevap

The rotation of a light paddle wheel placed directly in their path

Cathode rays consist of high-speed electrons possessing mass and kinetic energy. When these particles strike the vanes of a light paddle wheel inside a discharge tube, they transfer mechanical momentum to the wheel, causing it to roll along the glass rails. This directly proves that the rays carry kinetic energy and momentum.

Adım Adım Çözüm

1

Recall the physical properties of cathode rays evidenced by classical discharge tube experiments.

Cathode rays travel in straight lines, carry negative charge, produce fluorescence, and possess mechanical momentum.

Identifying the specific experimental evidence associated with kinetic energy and momentum.

2

Analyze the mechanical effect of particle bombardment on a physical object in a vacuum tube.

When stream particles strike the blades of a lightweight paddle wheel, they exert a mechanical force that causes the wheel to turn.

Transfer of momentum (

\Delta p

) upon collision converts particle kinetic energy (

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

) into rotational kinetic energy of the paddle wheel.

Anahtar Kavram

Particle Nature and Mechanical Energy of Cathode Rays

Tahmini Süre:45s

Soru 52Soru →

According to Bohr's model of the hydrogen atom, the total energy of an electron in the first excited state ( $n = 2$ ) is $-3.40\text{ eV}$ . Calculate the electric potential energy of the electron in this state, in electron-volts ( $\text{eV}$ ).

Cevabı ve açıklamayı göster

Cevap: -6.8

Cevap

The electric potential energy of the electron in the first excited state is

-6.80\text{ eV}

.

In Bohr's model of the hydrogen atom, an electron held in a circular orbit by electrostatic attraction has a kinetic energy

K = -E

and an electric potential energy

U = 2E

. Given a total energy

E = -3.40\text{ eV}

, multiplying by 2 yields

U = -6.80\text{ eV}

.

Adım Adım Çözüm

1

Identify the relationship between total energy and electric potential energy in the Bohr atomic model.

U = 2E

, where

U

is potential energy and

E

is total energy.

For an inverse-square electrostatic force binding an electron in a circular orbit, the virial theorem dictates that kinetic energy

K = -E

and potential energy

U = 2E

.

2

Substitute the given value of total energy

E = -3.40\text{ eV}

to solve for

U

.

U = 2 \times (-3.40\text{ eV}) = -6.80\text{ eV}

.

Direct multiplication yields the exact electric potential energy of the bound electron.

Anahtar Kavram

Energy components (kinetic, potential, and total energy) of an electron in Bohr's atomic model

Soru 53Soru →

In the study of electrical discharge through gases and cathode ray behavior, specific physical setups and pressure conditions produce distinct observable phenomena. Match each experimental condition or observation on the left with its corresponding underlying physical mechanism or property on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

1. Extension of the Crookes dark space matches with cessation of gas ionization causing direct glass fluorescence. 2. Maltese cross shadow matches with rectilinear propagation of cathode rays. 3. Magnetic field deflection into circular arc matches with centripetal magnetic Lorentz force (

F = qvB

). 4. Luminous striations match with periodic excitation, ionization, and recombination of gas molecules.

Each matching pair directly connects an observable discharge tube phenomenon with its fundamental physical principle: extreme evacuation (

0.01\text{ mmHg}

) allows unimpeded electron stream travel to fluoresce glass; obstacle shadows confirm rectilinear propagation; transverse magnetic fields induce circular motion via

evB

; and periodic energy exchange of electrons with gas molecules yields striations.

Adım Adım Çözüm

1

Analyze the pressure condition at

0.01\text{ mmHg}

At very low pressure (

0.01\text{ mmHg}

), gas collisions drop significantly, allowing cathode rays to reach the tube walls directly, extending the Crookes dark space throughout the tube and exciting glass fluorescence.

Mean free path increases beyond tube dimensions when gas density drops.

2

Analyze ray propagation using obstacle shadow formation

The sharp shadow cast by a Maltese cross demonstrates that cathode rays propagate in straight lines normal to the cathode.

Diffraction is negligible and ray trajectories do not bend around macroscopic obstacles.

3

Evaluate magnetic field interaction with cathode rays

The Lorentz force

F = q(\vec{v} \times \vec{B})

acts as a centripetal force (

evB = \frac{mv^2}{r}

), bending the negatively charged particle trajectory into a circle.

Moving electric charges experience magnetic forces perpendicular to velocity.

4

Identify the mechanism behind positive column striations

Striations represent repeating regions of inelastic electron collisions with gas atoms resulting in excitation and emission of light, followed by dark zones where electrons re-accelerate.

Quantized energy transfer during gas excitation creates spatial periodicity in luminescence.

Anahtar Kavram

Physical mechanisms of gaseous conduction across pressure stages and properties of cathode rays

Soru 54Soru →

An electron inside an excited atom drops from an energy state of $-1.20\text{ eV}$ to a lower energy state of $-4.50\text{ eV}$ . What is the frequency of the emitted electromagnetic radiation, in units of $10^{14}\text{ Hz}$ ? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The frequency of the emitted electromagnetic radiation is

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

(or 8.0 in units of

10^{14}\text{ Hz}

).

The energy of the photon emitted during a downward transition between discrete atomic energy levels is equal to the energy difference between the initial and final states. Calculating

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

, converting to Joules gives

3.30 \times 1.6 \times 10^{-19}\text{ J} = 5.28 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant

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

yields a frequency of

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

.

Adım Adım Çözüm

1

Determine the energy of the emitted photon in electron-volts

\Delta E = 3.30\text{ eV}

The energy of the emitted photon equals the difference between the upper and lower atomic energy levels:

\Delta E = -1.20\text{ eV} - (-4.50\text{ eV}) = 3.30\text{ eV}

.

2

Convert photon energy from electron-volts to Joules

\Delta E = 5.28 \times 10^{-19}\text{ J}

Since Planck's constant is given in SI units (J s), energy must be converted to Joules using

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

.

3

Calculate photon frequency using Planck's relation

f = 8.0 \times 10^{14}\text{ Hz}

Applying

f = \frac{\Delta E}{h} = \frac{5.28 \times 10^{-19}\text{ J}}{6.6 \times 10^{-34}\text{ J s}}

yields

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

.

Anahtar Kavram

Photon Emission and Energy Level Transitions

Soru 55Soru →

A hydrogen atom initially in its ground state with energy $E_1 = -13.6\text{ eV}$ absorbs a photon with energy $12.75\text{ eV}$ , raising the electron to an excited quantum level $n$ . The atom then undergoes de-excitation to lower energy levels. Taking Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , speed of light $c = 3.0 \times 10^8\text{ m/s}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ , what is the shortest wavelength of light emitted during any of the possible downward transitions?

9.7 \times 10^{-8}\text{ m}

1.55 \times 10^{-26}\text{ m}

1.88 \times 10^{-6}\text{ m}

1.02 \times 10^{-7}\text{ m}

Cevabı ve açıklamayı göster

Cevap: $9.7 \times 10^{-8}\text{ m}$

Cevap

9.7 \times 10^{-8}\text{ m}

The correct answer of

9.7 \times 10^{-8}\text{ m}

is obtained by finding that the absorbed photon of

12.75\text{ eV}

promotes the electron to the

n = 4

state (

E_4 = -0.85\text{ eV}

). The shortest wavelength photon is emitted in the transition with the greatest energy difference, which is

n = 4 \rightarrow n = 1

(

\Delta E = 12.75\text{ eV}

). Converting

12.75\text{ eV}

to

2.04 \times 10^{-18}\text{ J}

and substituting into

\lambda = \frac{hc}{\Delta E}

yields

9.7 \times 10^{-8}\text{ m}

.

Adım Adım Çözüm

1

Determine the energy of the excited state

E_n

E_n = E_1 + \Delta E_{\text{absorbed}} = -13.6\text{ eV} + 12.75\text{ eV} = -0.85\text{ eV}

Absorbing energy elevates the atom from its ground state energy to a higher energy level.

2

Identify the principal quantum number

n

of the excited state

n = \sqrt{\frac{-13.6\text{ eV}}{-0.85\text{ eV}}} = \sqrt{16} = 4

For hydrogen,

E_n = \frac{E_1}{n^2}

.

3

Identify the transition giving the shortest wavelength

Transition from

n = 4 \rightarrow n = 1

with maximum energy change

\Delta E_{\text{max}} = 12.75\text{ eV}

Since

\lambda = \frac{hc}{\Delta E}

, the shortest wavelength occurs at maximum energy emission.

4

Convert energy to Joules and calculate wavelength

\Delta E = 12.75 \times 1.6 \times 10^{-19}\text{ J} = 2.04 \times 10^{-18}\text{ J}

;

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{2.04 \times 10^{-18}} \approx 9.7 \times 10^{-8}\text{ m}

Applying the photon energy-wavelength relation

E = \frac{hc}{\lambda}

in SI units.

Anahtar Kavram

Energy Levels and Atomic Spectra

Tahmini Süre:3m 0s

Soru 56Soru →

An electron in an excited state of an atom moves from an energy level of $-2.80\text{ eV}$ to a lower energy state of $-7.60\text{ eV}$ . What is the energy of the emitted photon in Joules? ( $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

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

4.80\text{ J}

1.66 \times 10^{-18}\text{ J}

1.22 \times 10^{-18}\text{ J}

Cevabı ve açıklamayı göster

Cevap: $7.68 \times 10^{-19}\text{ J}$

Cevap

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

The emitted photon's energy is given by the difference between the initial higher state and the final lower state:

\Delta E = E_2 - E_1 = -2.80\text{ eV} - (-7.60\text{ eV}) = 4.80\text{ eV}

. Converting to Joules gives

4.80 \times 1.6 \times 10^{-19}\text{ J} = 7.68 \times 10^{-19}\text{ J}

.

Adım Adım Çözüm

1

Calculate the energy difference between the initial and final states in electron-volts (eV)

\Delta E = E_{\text{initial}} - E_{\text{final}} = -2.80\text{ eV} - (-7.60\text{ eV}) = 4.80\text{ eV}

The energy of an emitted photon during a downward atomic transition equals the difference in energy between the two states.

2

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

E = 4.80 \times 1.6 \times 10^{-19}\text{ J} = 7.68 \times 10^{-19}\text{ J}

SI units require energy to be expressed in Joules, where

1\text{ eV} = 1.6 \times 10^{-19}\text{ J}

.

Anahtar Kavram

Energy level transitions and photon emission

Soru 57Soru →

In a cathode-ray experiment, electrons are accelerated from rest through a potential difference of $100\text{ V}$ . Taking the specific charge ( $\frac{e}{m}$ ) of an electron to be $1.80 \times 10^{11}\text{ C/kg}$ , calculate the final speed of the electrons as they pass through the anode aperture, expressed in units of $10^6\text{ m/s}$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The final speed of the electrons is

6.0 \times 10^6\text{ m/s}

, which corresponds to a coefficient value of 6.0.

By applying conservation of energy, the kinetic energy acquired by electrons in a cathode-ray tube equals the electric work done on them (

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

). Solving for velocity gives

v = \sqrt{2V(e/m)}

. Substituting

V = 100\text{ V}

and

e/m = 1.80 \times 10^{11}\text{ C/kg}

gives

v = 6.0 \times 10^6\text{ m/s}

.

Adım Adım Çözüm

1

Set up the energy conservation relation for cathode ray electrons.

Electrical work done

W = eV

equals kinetic energy

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

.

Electrons starting from rest gain kinetic energy equal to the electrical potential energy lost across the potential difference.

2

Rearrange the formula to isolate the electron velocity

v

.

v^2 = 2V\left(\frac{e}{m}\right) \implies v = \sqrt{2V\left(\frac{e}{m}\right)}

.

Isolating velocity allows direct calculation using the given specific charge

(\frac{e}{m})

and accelerating voltage

V

.

3

Substitute given values into the equation and compute

v

.

v = \sqrt{2 \times 100 \times 1.80 \times 10^{11}} = \sqrt{36 \times 10^{12}} = 6.0 \times 10^6\text{ m/s}

.

Evaluating the square root yields the speed in meters per second.

Anahtar Kavram

Electron acceleration in cathode ray tubes and specific charge relation

Soru 58Soru →

In discharge tube experiments, electrical conduction in gases transitions through distinct physical regimes as the internal gas pressure is progressively reduced. Match each discharge phenomenon with its corresponding physical cause or operational pressure condition.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The correct pairings are: (1) Formation of the luminous positive column matches continuous de-excitation and radiative recombination of gas atoms at pressures around

1.0\text{ mmHg}

; (2) Appearance and expansion of the Crookes dark space matches increase in the electron mean free path at pressures around

0.01\text{ mmHg}

; (3) Emission of high-velocity cathode rays matches bombardment of the cathode by energetic positive ions releasing secondary electrons below

0.01\text{ mmHg}

; and (4) Complete cessation of electric current flow matches extensive evacuation below

10^{-4}\text{ mmHg}

leaving insufficient gas molecules for ionization.

Conduction through gases relies heavily on pressure. At moderate low pressure (

1.0\text{ mmHg}

), excited gas atoms emit light forming the positive column. Decreasing pressure to

0.01\text{ mmHg}

increases the electron mean free path to produce the Crookes dark space and generates energetic cathode rays through ion bombardment. Extreme evacuation (

< 10^{-4}\text{ mmHg}

) removes all gas charge carriers, halting electric conduction.

Adım Adım Çözüm

1

Analyze the pressure regime of

1.0\text{ mmHg}

in a discharge tube.

Identified the positive column as the main luminous region filling most of the tube due to atom excitation and light emission.

At this pressure, gas density is sufficient to undergo repeated inelastic collisions that excite atoms and produce visible glow.

2

Examine the physical origin of the Crookes dark space at

0.01\text{ mmHg}

.

Understood that lower gas density increases electron mean free path.

Electrons near the cathode travel a longer distance before hitting gas particles, creating a dark gap where collisions do not occur.

3

Determine how cathode rays are emitted at very low pressures.

Linked cathode ray emission to positive ion impact on the cathode surface.

High electric fields accelerate residual positive ions to strike the cathode, causing secondary electron emission.

4

Evaluate the extreme vacuum limit below

10^{-4}\text{ mmHg}

.

Concluded that current stops when gas particles are virtually absent.

Gases conduct electricity via ion and electron production from collision ionization; eliminating gas molecules prevents charge transport.

Anahtar Kavram

Pressure-dependent regimes of gas conduction and cathode ray generation

Tahmini Süre:2m 0s

Soru 59Soru →

An electron inside an excited atom drops from an energy state of $-2.10\text{ eV}$ to a lower energy state of $-4.575\text{ eV}$ . What is the wavelength of the emitted photon in nanometers ( $\text{nm}$ )? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^8\text{ m/s}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ ).

Cevabı ve açıklamayı göster

Cevap: 500

Cevap

The wavelength of the emitted photon is

500\text{ nm}

.

The energy of the emitted photon is calculated from the energy change of the electron,

\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}

. Converting this to Joules gives

2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}

. Substituting into the wavelength formula

\lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.96 \times 10^{-19}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}

.

Adım Adım Çözüm

1

Calculate the energy difference between the initial excited state and final state

\Delta E = -2.10\text{ eV} - (-4.575\text{ eV}) = 2.475\text{ eV}

The energy of the emitted photon equals the difference between the two energy levels.

2

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

\Delta E = 2.475 \times 1.6 \times 10^{-19}\text{ J} = 3.96 \times 10^{-19}\text{ J}

Standard SI units must be used to calculate wavelength in meters.

3

Apply the Planck-Einstein relation

\lambda = \frac{hc}{\Delta E}

to calculate the wavelength in meters and convert to nanometers

\lambda = \frac{(6.6 \times 10^{-34}\text{ J s})(3.0 \times 10^8\text{ m/s})}{3.96 \times 10^{-19}\text{ J}} = 5.0 \times 10^{-7}\text{ m} = 500\text{ nm}

Converting meters to nanometers requires multiplying by

10^9

.

Anahtar Kavram

Photon emission wavelength during atomic energy level transitions

Tahmini Süre:1m 30s

Soru 60Soru →

X-rays are high-frequency electromagnetic waves that undergo deflection when passing through strong electric or magnetic fields.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is False because X-rays carry no electrical charge and therefore pass through electric and magnetic fields without deflection.

X-rays are electromagnetic radiation made of photons with zero electric charge. Because they carry no charge, they travel in straight lines and remain entirely unaffected by surrounding electric or magnetic fields.

Adım Adım Çözüm

1

Identify the physical nature of X-rays.

X-rays are electromagnetic radiation (photons) of high frequency and short wavelength.

Electromagnetic waves travel at the speed of light and consist of transverse oscillating electric and magnetic fields, carrying energy but no electric charge.

2

Analyze the effect of electric and magnetic fields on uncharged radiation.

The electric force

F_E = qE

and magnetic force

F_B = qvB sin\theta

are both zero when

q = 0

.

Since photons have zero net electrical charge (

q = 0

), they experience no net deflecting force in external electric or magnetic fields.

Anahtar Kavram

Neutrality of X-rays in electromagnetic fields

Atomic and Nuclear Physics

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