PHYSM-solve 4 5 6Ruofstion 4A photocell is sketche

solve 4 5 6

Solution

a

The key measurement for determining the maximum energy of emitted photoelectrons in a photocell is the smallest voltage to just achieve minimum current. This is because the stopping potential (smallest voltage to stop the current) directly relates to the maximum kinetic energy of the emitted photoelectrons

4 Answer

D

Key Concept

Photoelectric Effect: The photoelectric effect occurs when light shines on a material and ejects electrons from it. The maximum kinetic energy of these electrons is given by the stopping potential. Equation:

KE_{\text{max}} = eV_{\text{stop}}

(where

e

is the electron charge and

V_{\text{stop}}

is the stopping potential)

Explanation

The stopping potential is the smallest voltage needed to stop the photoelectrons, which directly measures their maximum kinetic energy.

Solution

a

Einstein's explanation of the photoelectric effect introduced the idea that the energy carried by light is in discrete quantities, or quanta. This was a significant insight that led to the development of quantum mechanics

5 Answer

D

Key Concept

Quantum Theory: Einstein proposed that light consists of photons, each carrying a quantum of energy proportional to its frequency. Equation:

E = hf

(where

E

is the energy,

h

is Planck's constant, and

f

is the frequency of the light)

Explanation

Einstein's insight was that light energy is quantized, meaning it comes in discrete packets called photons. This was crucial for explaining the photoelectric effect.

Solution

a

A dark region in a two-slit interference pattern occurs due to destructive interference, where the path difference between the two waves is an odd multiple of half wavelengths, causing them to cancel each other out

6 Answer

D

Key Concept

Interference of Light: In a two-slit experiment, light waves interfere constructively (bright regions) or destructively (dark regions) depending on their path difference. Equation:

\Delta x = (m + \frac{1}{2})\lambda

for destructive interference (where

m

is an integer and

\lambda

is the wavelength)

Explanation

Destructive interference causes dark regions because the waves cancel each other out, resulting in no detectable photons at those points.

solve 7 8 9 10

Solution

a

When the intensity of the source of light for the interference pattern is reduced greatly, the number of photons is reduced greatly. This is because intensity is directly proportional to the number of photons

Answer

A

Key Concept

Intensity of Light: The intensity of light is directly proportional to the number of photons. When the intensity is reduced, the number of photons is also reduced.

Explanation

Reducing the intensity of light means fewer photons are emitted, which directly affects the interference pattern.

Solution

a

Photoelectrons are emitted almost immediately from the metal surface after light of a suitable wavelength strikes it because only one photon is required to liberate one electron. This is a fundamental principle of the photoelectric effect

Answer

A

Key Concept

Photoelectric Effect: The photoelectric effect occurs when photons of sufficient energy strike a metal surface and liberate electrons. Each photon can liberate one electron if its energy is greater than the work function of the metal.

Explanation

The immediate emission of photoelectrons is explained by the fact that a single photon can liberate a single electron if it has enough energy.

Solution

a

When the light intensity is increased but the frequency remains the same, the number of photoelectrons emitted increases while the maximum kinetic energy of the photoelectrons remains the same. This is because the energy of each photon (and thus the maximum kinetic energy of the photoelectrons) depends on the frequency, not the intensity

Answer

C

Key Concept

Photoelectric Effect: The number of photoelectrons emitted is proportional to the intensity of the light, while the maximum kinetic energy of the photoelectrons depends on the frequency of the light.

Explanation

Increasing the intensity increases the number of photons, and thus the number of photoelectrons, but does not change the energy of each photon.

Solution

a

The minimum energy of an incoming photon required to eject a photoelectron is equal to the work function of the metal. Given the work function is

2.0 \, \text{eV}

, we convert this to joules:

2.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-19} \, \text{J}

Answer

A

Key Concept

Work Function: The work function is the minimum energy required to eject an electron from a metal surface. It is often given in electron volts (eV) and can be converted to joules (J).

Explanation

The work function represents the minimum energy needed for a photon to eject an electron, and this energy can be converted from eV to J for calculation purposes.

solve 11

Solution

a

The graph shows the maximum kinetic energy (

E_{K \max}

) of photoelectrons versus the frequency (

f

) of light. The work function (

\phi

) of the metal is the minimum energy required to eject an electron from the metal surface. This is given by the equation:

E_{K \max} = hf - \phi

where

h

is Planck's constant and

f

is the frequency of the incident light

b

When the work function of the metal increases, the threshold frequency (

f_0

) also increases. The threshold frequency is given by:

f_0 = \frac{\phi}{h}

Thus, for a larger work function, the new threshold frequency will be higher

c

On the graph, this means that the line representing the relationship between

E_{K \max}

and

f

will shift to the right. The slope of the line, which is

h

, remains unchanged, but the intercept on the frequency axis (where

E_{K \max} = 0

) will be at a higher frequency

Answer

The graph with the dashed line shifted to the right, indicating a higher threshold frequency, represents the results from the second experiment.

Key Concept

The photoelectric effect and the relationship between the maximum kinetic energy of photoelectrons, the frequency of incident light, and the work function of the metal.

Explanation

When the work function of the metal increases, the threshold frequency required to emit photoelectrons also increases, resulting in a rightward shift of the graph.

solve 11 with process and answer

Solution

a

The photoelectric effect is described by the equation:

E_k = hf - \phi

where

E_k

is the maximum kinetic energy of the photoelectrons,

h

is Planck's constant,

f

is the frequency of the incident light, and

\phi

is the work function of the metal

b

In the graph, the y-intercept represents the negative of the work function

-\phi

. A larger work function means a more negative y-intercept

c

The slope of the line, which is

h

, remains unchanged because Planck's constant is a universal constant

d

Therefore, the graph for the second experiment with a larger work function will have the same slope but a lower y-intercept. This corresponds to a parallel line shifted downwards

Answer

Graph with a parallel line shifted downwards

Key Concept

The photoelectric effect and its relationship with the work function and frequency of incident light.

Explanation

A larger work function results in a more negative y-intercept in the graph of maximum kinetic energy versus frequency, while the slope remains the same.

solve 12 13

Solution

a

The de Broglie wavelength of an electron is given by

\lambda = \frac{h}{p}

, where

h

is Planck's constant and

p

is the momentum of the electron

b

For an electron to form a stable orbit, the circumference of the orbit must be an integer multiple of the de Broglie wavelength:

2\pi r = n\lambda

, where

r

is the radius of the orbit and

n

is an integer

c

Given

\lambda = 314 \, \text{pm}

and

r = 100 \, \text{pm}

, we can check if the circumference

2\pi r

is an integer multiple of

\lambda

:

2\pi \times 100 \, \text{pm} \approx 628 \, \text{pm}

Since

628 \, \text{pm}

is approximately twice

314 \, \text{pm}

, it suggests that

n = 2

d

Therefore, the electron can form a stable orbit with a radius of

100 \, \text{pm}

because the circumference is an integer multiple of the de Broglie wavelength

12 Answer

Yes, the electron can form a stable orbit with a radius of 100 pm.

Key Concept

The de Broglie wavelength and quantization of orbits: For an electron to form a stable orbit, the circumference of the orbit must be an integer multiple of the de Broglie wavelength.

Explanation

By checking if the circumference of the orbit is an integer multiple of the de Broglie wavelength, we can determine if the orbit is allowed.

Question 13

a

The energy of a photon is related to its wavelength by the equation:

E = \frac{hc}{\lambda}

, where

h

is Planck's constant and

c

is the speed of light

b

Given the energy transition

E = 5.9 \, \text{eV}

, we need to convert this energy to joules:

1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}

E = 5.9 \times 1.602 \times 10^{-19} \, \text{J} = 9.4518 \times 10^{-19} \, \text{J}

c

Using the equation

\lambda = \frac{hc}{E}

, where

h = 6.626 \times 10^{-34} \, \text{Js}

and

c = 3 \times 10^8 \, \text{m/s}

:

\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{9.4518 \times 10^{-19}} \approx 2.1 \times 10^{-7} \, \text{m} = 210 \, \text{nm}

13 Answer

The wavelength of the photon is approximately 210 nm.

Key Concept

The relationship between energy and wavelength of a photon:

E = \frac{hc}{\lambda}

.

Explanation

By converting the energy transition from eV to joules and using the equation

\lambda = \frac{hc}{E}

, we can calculate the wavelength of the resulting photon.

solve 14 15 16

Solution

a

Use the de Broglie wavelength formula:

\lambda = \frac{h}{mv}

where

h

is Planck's constant (

6.626 \times 10^{-34} \, \text{Js}

),

m

is the mass of the electron (

9.11 \times 10^{-31} \, \text{kg}

), and

v

is the velocity of the electron (

2.0 \times 10^6 \, \text{m/s}

).

\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 2.0 \times 10^6}

\lambda = 3.63 \times 10^{-10} \, \text{m}

Convert to nanometers:

\lambda = 0.363 \, \text{nm}

b

The wave nature of electrons explains quantized energy levels in atoms through the concept of standing waves. Electrons in an atom can only occupy certain orbits where their de Broglie wavelength forms a standing wave. This means that only certain discrete wavelengths (and thus energies) are allowed, leading to quantized energy levels

Question 15

a

The energy level diagram for a sodium atom shows transitions between different energy states. To find the energy of a photon emitted during a transition, use the energy difference between the initial and final states. For example, if an electron transitions from -3.61 eV to -10.0 eV:

\Delta E = E_{\text{final}} - E_{\text{initial}}

\Delta E = -10.0 \, \text{eV} - (-3.61 \, \text{eV})

\Delta E = -6.39 \, \text{eV}

The negative sign indicates energy is released

Question 16

a

To calculate the frequency of the emitted photon, use the energy difference and Planck's equation:

E = h \nu

where

E

is the energy difference (

6.39 \, \text{eV}

),

h

is Planck's constant (

4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s}

), and

\nu

is the frequency.

\nu = \frac{E}{h}

\nu = \frac{6.39}{4.135667696 \times 10^{-15}}

\nu = 1.54 \times 10^{15} \, \text{Hz}

Answer

14a: 0.363 nm

14b: Quantized energy levels are explained by the standing wave nature of electron de Broglie wavelengths.

15a: -6.39 eV

16a: 1.54 × 10^15 Hz

Key Concept

de Broglie wavelength, quantized energy levels, energy transitions in atoms, Planck's equation

Explanation

The de Broglie wavelength formula helps calculate the wavelength of an electron in motion. The wave nature of electrons leads to quantized energy levels due to standing wave formation. Energy transitions in atoms involve discrete energy changes, which can be calculated and related to photon frequency using Planck's equation.

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