Question

The straight line graphs obtained for two metals:

• A. coincide each other.
• B. are parallel to each other.
• C. are not parallel to each other and cross at a point on \( \nu \)-axis.
• D. are not parallel to each other and do not cross at a point on \( \nu \)-axis.
• A. \( \frac{e}{m} \)
• B. \( \frac{1}{m} \)
• C. \( \frac{me}{e} \)
• D. \( \frac{m}{e} \)
• A. \( \frac{h \nu_0}{e} \), \( V_0 \)
• B. \( \nu_0 \), \( h \nu_0 \)
• C. \( \frac{h \nu_0}{e} \), \( eV_0 \)
• D. \( h \nu_0 \), \( h \nu_0 \)
• A. \(2,\frac{1}{2}\)
• B. \(\frac{1}{2},\frac{1}{2}\)
• C. \(\frac{1}{2},2\)
• D. 2,2
• A. \(1.33 \mu m\)
• B. \(3.3 \mu m\)
• C. \(16.6 \mu m\)
• D. \(13.3 \mu m\)

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the relationship between the stopping potential \(V_0\) and the frequency \(\nu\) for two different metals under the photoelectric effect. According to the given information, the maximum kinetic energy of photoelectrons is represented as:

\[K_m = eV_0 = h(\nu - \nu_0)\]

Here, \(e\) is the charge of an electron, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident photon, and \(\nu_0\) is the threshold frequency of the metal.

The equation \(eV_0 = h(\nu - \nu_0)\) can be rewritten as:

\[V_0 = \frac{h}{e} \nu - \frac{h}{e} \nu_0\]

This is a straight line equation \(y = mx + c\), where:

• \(y\) is \(V_0\)
• \(x\) is \(\nu\)
• \(m = \frac{h}{e}\) is the slope
• \(c = -\frac{h}{e} \nu_0\) is the intercept

The slope \(m\) is determined by \(\frac{h}{e}\), which is constant for both metals since Planck's constant and the electron charge are universal constants. Therefore, the straight-line graphs of \(V_0\) versus \(\nu\) for two different metals will have the same slope, meaning:

The lines are parallel to each other.

Answer 2

To determine the value of Planck's constant for a given metal, we need to analyze the relationship between the stopping potential \( V_0 \) and the frequency \( \nu \) of the incident photons. The equation governing the photoelectric effect is:

\( eV_0 = h(\nu - \nu_0) \)

Here, \( e \) is the elementary charge, \( h \) is Planck's constant, \( \nu_0 \) is the threshold frequency, and \( K_m \) is the maximum kinetic energy of the photoelectrons.

From this equation, if we rearrange it in the form of a straight line, we obtain:

\( V_0 = \frac{h}{e}\nu - \frac{h}{e}\nu_0 \)

Comparing this with the standard line equation \( y = mx + c \), we can see:

\( m = \frac{h}{e} \)

Thus, the slope \( m \) of the graph between the stopping potential \( V_0 \) and the frequency \( \nu \) is equal to \( \frac{h}{e} \). So, to find the value of Planck's constant \( h \) for the metal, we rearrange for \( h \):

\( h = me \)

Given the slope \( m \), we observe from the options that the solution to express \( h \) as a function of known quantities is:

\( h = \frac{e}{m} \)

Therefore, the correct option for Planck's constant is: \( \frac{e}{m} \)

Answer 3

The photoelectric effect describes the emission of electrons from a metal surface when illuminated by light of a certain frequency. The energy required to release an electron is related to the frequency of the incident light. The kinetic energy \(K_m\) of the emitted electron can be expressed by the equation:

\(K_m=h(\nu-\nu_0)\)

where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\nu_0\) is the threshold frequency of the metal.

The stopping potential \(V_0\) is the potential needed to stop the fastest electrons, hence it is related to the maximum kinetic energy:

\(eV_0=h(\nu-\nu_0)\)

This equation represents a straight-line graph when plotted, where \(V_0\) is the dependent variable with \(\nu\) as the independent variable. The slope of the line is \(h/e\), and the y-intercept is where \(V_0=0\) when \(\nu=\nu_0\). Solving for this gives:

\(0=h(\nu_0-\nu_0)\)

The x-intercept is \(\nu=\nu_0\) and the y-intercept (value of \(V_0\)) is when:

\(V_0=\frac{h\nu_0}{e}\)

Hence, the intercepts on the \(\nu\)-axis (x-axis) and \(V_0\)-axis (y-axis) are \(\frac{h\nu_0}{e}\) and \(V_0\), respectively, matching the option: \(\frac{h\nu_0}{e}\), \(V_0\).

Answer 4

To determine how the wave number and frequency of a photon change when its wavelength is doubled, we follow the equations that relate these quantities to the wavelength. The wave number \(k\) is defined as the reciprocal of the wavelength \(\lambda\): 
\(k = \frac{1}{\lambda}\)
When the wavelength is doubled, \(\lambda\) becomes \(2\lambda\). The new wave number \(k'\) is:
\(k' = \frac{1}{2\lambda} = \frac{1}{2}k\)
This means the wave number is halved.
The frequency \(f\) is related to the wavelength by the speed of light \(c\):
\(c = \lambda f\)
When wavelength is \(2\lambda\), the new frequency \(f'\) is calculated from:
\(c = (2\lambda)f'\)
So,
\(f' = \frac{c}{2\lambda} = \frac{1}{2}f\)
This shows the frequency is also halved.
Thus, when the wavelength is doubled, the wave number and frequency both become half of their original values. Therefore, the correct answer is \(\frac{1}{2},\frac{1}{2}\).

Answer 5

The momentum \( p \) of a photon is related to its wavelength \( \lambda \) by the equation: \[ p = \frac{h}{\lambda} \] Where \( h \) is Planck's constant. Rearranging for \( \lambda \): \[ \lambda = \frac{h}{p} \] Substituting the known values: - \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), - \( p = 5.0 \times 10^{-29} \, \text{kg} \cdot \text{m/s} \). \[ \lambda = \frac{6.626 \times 10^{-34}}{5.0 \times 10^{-29}} = 1.33 \times 10^{-5} \, \text{m} = 13.3 \, \mu\text{m} \] Thus, the wavelength of the photon is: % Correct Answer Correct Answer:} (D) 13.3 m