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

The graph between \( V_0 \) and \( \nu \) for two metals shows that both graphs are straight lines, indicating a linear relationship between the stopping potential and frequency. Since the slope of the graph is constant, the lines are parallel to each other. Thus, the correct answer is , as the straight line graphs obtained for two metals are parallel to each other.

Answer 2

From the equation \( eV_0 = h(\nu - \nu_0) \), comparing it with the equation of a straight line \( y = mx + c \), the slope \( m \) is given by: \[ m = \frac{h}{e} \] This shows that Planck’s constant \( h \) can be expressed as \( h = e \times m \), where \( m \) is the slope of the graph and \( e \) is the charge of the electron. Thus, the correct value of Planck's constant for this metal is \( \frac{e}{m} \).

Answer 3

The intercept on the \( V_0 \)-axis is \( V_0 \) when \( \nu = \nu_0 \), and the intercept on the \( \nu \)-axis occurs when \( V_0 = 0 \). Thus: - The intercept on the \( V_0 \)-axis gives the stopping potential \( V_0 \) corresponding to \( \nu = \nu_0 \). - The intercept on the \( \nu \)-axis gives \( \nu_0 \), the threshold frequency. Thus, the intercepts on the axes are \( \frac{h \nu_0}{e} \) for the \( V_0 \)-axis and \( \nu_0 \) for the \( \nu \)-axis.

Answer 4

The wavelength \( \lambda \) and frequency \( \nu \) of a photon are related by the equation: \[ c = \lambda \nu \] Where: - \( c \) is the speed of light, - \( \lambda \) is the wavelength, - \( \nu \) is the frequency. When the wavelength \( \lambda \) is doubled, the frequency \( \nu \) becomes halved, because the speed of light \( c \) is constant. Therefore: \[ \nu' = \frac{\nu}{2} \] The wave number \( k \), which is the reciprocal of the wavelength, is given by: \[ k = \frac{1}{\lambda} \] When the wavelength is doubled, the wave number becomes halved: \[ k' = \frac{k}{2} \] Thus, the wave number becomes \( \frac{1}{2} \) times, and the frequency becomes \( \frac{1}{2} \) times. Thus, the correct answer is: % Correct Answer Correct Answer:} \( \frac{1}{2} \), \( \frac{1}{2} \)

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