Question

When the energy of the incident radiation is increased by $20 \%, $ the kinetic energy of the photoelectrons emitted from a metal surface increased from 0.5 eV to 0.8 eV. The work function of the metal is

• A. 0.65 eV
• B. 1.0 eV
• C. 1.3 eV
• D. 1.5 eV

Answer 1

The Correct Option is B

The problem involves calculating the work function of a metal using the photoelectric effect, which is described by the equation:

K.E. = h\nu - \phi

where:

• K.E. is the kinetic energy of the emitted photoelectrons.
• h\nu is the energy of the incident radiation.
• \phi is the work function of the metal.

Initially, the kinetic energy of the photoelectrons is 0.5 eV. When the energy of the incident radiation is increased by 20%, the kinetic energy increases to 0.8 eV. Let's denote:

• h\nu = E
• h\nu' = 1.2E (after the 20% increase)

Initially, from the photoelectric equation:

0.5 \, \text{eV} = E - \phi

After increasing the energy by 20%, the equation becomes:

0.8 \, \text{eV} = 1.2E - \phi

We now have two equations:

1. E - \phi = 0.5 \, \text{eV}
2. 1.2E - \phi = 0.8 \, \text{eV}

Subtract equation (1) from equation (2) to eliminate \phi:

(1.2E - \phi) - (E - \phi) = 0.8 - 0.5

0.2E = 0.3 \, \text{eV}

Solve for E:

E = \frac{0.3}{0.2} = 1.5 \, \text{eV}

Substitute E back into the first equation:

1.5 \, \text{eV} - \phi = 0.5 \, \text{eV}

Solve for \phi:

\phi = 1.5 \, \text{eV} - 0.5 \, \text{eV} = 1.0 \, \text{eV}

Thus, the work function of the metal is 1.0 eV.