Question

The surface of a metal is first illuminated with a light of wavelength 300 nm and later illuminated by another light of wavelength 500 nm. It is observed that the ratio of maximum velocities of photoelectrons in two cases is 3. The work function of the metal value is close to:

• A. \( 6.48 \text{ eV} \)
• B. \( 1.23 \text{ eV} \)
• C. \( 4.17 \text{ eV} \)
• D. \( 2.28 \text{ eV} \)

Hint:

To determine the work function of a metal, use the photoelectric equation and known photon energy values.

Answer 1

The Correct Option is D

The problem involves the photoelectric effect, where the maximum kinetic energy of photoelectrons is given by the equation:

\[ KE_{\text{max}} = h\nu - \phi \]

where \( KE_{\text{max}} \) is the maximum kinetic energy, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident light, and \( \phi \) is the work function of the metal.

The frequency \( \nu \) can be expressed in terms of the speed of light \( c \) and the wavelength \( \lambda \) as \( \nu = \frac{c}{\lambda} \). Therefore, the kinetic energy becomes:

\[ KE_{\text{max}} = \frac{hc}{\lambda} - \phi \]

Given the wavelengths \( \lambda_1 = 300 \, \text{nm} \) and \( \lambda_2 = 500 \, \text{nm} \), and the ratio of maximum velocities \( \frac{v_1}{v_2} = 3 \), the kinetic energy is related to velocity as \( KE_{\text{max}} = \frac{1}{2}mv^2 \), where \( m \) is the mass of an electron.

Let's denote the photoelectric equation for both cases:

\[ \frac{hc}{300 \times 10^{-9}} - \phi = \frac{1}{2}m(3v_2)^2 \]

\[ \frac{hc}{500 \times 10^{-9}} - \phi = \frac{1}{2}mv_2^2 \]

Taking the ratio of the kinetic energy equations and solving for \( \phi \):

\[ \frac{\frac{hc}{300 \times 10^{-9}} - \phi}{\frac{hc}{500 \times 10^{-9}} - \phi} = \left(\frac{3v_2}{v_2}\right)^2 = 9 \]

Substituting the constant values \( h = 6.63 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \):

Simplify using \( hc = 1240 \, \text{eV}\cdot \text{nm} \):

\[ \frac{1240/300 - \phi}{1240/500 - \phi} = 9 \]

\[ \frac{4.133 - \phi}{2.48 - \phi} = 9 \]

Solving for \( \phi \):

\[ 4.133 - \phi = 9(2.48 - \phi) \]

\[ 4.133 - \phi = 22.32 - 9\phi \]

\[ 8\phi = 22.32 - 4.133 \]

\[ 8\phi = 18.187 \]

\[ \phi = \frac{18.187}{8} \approx 2.27 \, \text{eV} \]

The closest option to this calculated value is

\( 2.28 \text{ eV} \).

Answer 2

Step 1: The energy of incident photons is given by the equation: \[ E = \frac{hc}{\lambda} \] where \( h = 6.63 \times 10^{-34} \) J.s, \( c = 3 \times 10^8 \) m/s, and \( \lambda \) is the wavelength.
Step 2: Compute photon energies: \[ E_1 = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{300 \times 10^{-9}} \] \[ E_2 = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{500 \times 10^{-9}} \]
Step 3: Use the given velocity ratio: \[ \frac{v_1}{v_2} = 3 \]
Step 4: Using Einstein's photoelectric equation: \[ KE = E - \phi \] By solving for \( \phi \), we get: \[ \phi \approx 2.28 \text{ eV} \]