Question

If the frequency of the incident light on a metal surface is increased by 10%, then the kinetic energy of the emitted photoelectrons is increased from 0.5 eV to 0.7 eV. Then the work function of the metal is:

• A. 1 eV
• B. 1.2 eV
• C. 1.5 eV
• D. 1.8 eV
• E. 2 eV

Hint:

In photoelectric effect problems, the change in kinetic energy of photoelectrons can be used to find the work function by applying the photoelectric equation and solving for \( \phi \).

Answer 1

The Correct Option is C

We know that the kinetic energy of the emitted photoelectrons is given by the photoelectric equation: \[ K.E. = h f - \phi \] where: - \( K.E. \) is the kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal.
Let the initial frequency be \( f_1 \) and the corresponding kinetic energy be \( K.E_1 = 0.5 \, {eV} \): \[ K.E_1 = h f_1 - \phi \tag{1} \] Let the final frequency be \( f_2 = 1.1 f_1 \) (since the frequency is increased by 10%), and the corresponding kinetic energy be \( K.E_2 = 0.7 \, {eV} \): \[ K.E_2 = h f_2 - \phi = h (1.1 f_1) - \phi \tag{2} \] Now, subtract equation (1) from equation (2): \[ K.E_2 - K.E_1 = h (1.1 f_1) - \phi - (h f_1 - \phi) \] \[ 0.7 - 0.5 = h (1.1 f_1 - f_1) \] \[ 0.2 = h \times 0.1 f_1 \] \[ h f_1 = 2 \, {eV} \] Substitute this value into equation (1): \[ 0.5 = 2 - \phi \] \[ \phi = 1.5 \, {eV} \] Thus, the work function of the metal is 1.5 eV, which corresponds to option (C).