Question

When the energy of the incident radiation is increased by 20%, The kinetic energy of the photelectrons emitted from a metal surfce in creased from 0.5V to 0.8eV The work function of the metal is

• A. 0.65 eV
• B. 1.0 eV
• C. 1.2 eV
• D. 1.5 eV

Answer 1

The correct option is B 1.0 eV

Given: $h{\nu }_2=1.2h{\nu }_1$ $.....\left(1\right)$

$K{E}_1=0.5eV$

$K{E}_2=0.8eV$

To find: Work function,$W_0=?$

Solution: As we know that,

$KE=h\nu -W_0$

Case 1: $K{E}_1=h{\nu }_1-W_0$

$==>h{\nu }_1=K{E}_1+W_0$

$==>h{\nu }_1=0.5+W_0$ $.....\left(2\right)$

Case 2: $K{E}_2=h{\nu }_2-W_0$

$==>W_0=h{\nu }_2-0.8$

On using $e{q}^n\left(1\right)$

$==>W_0=1.2h{\nu }_1-0.8$

On using $e{q}^n\left(2\right)$

$==>W_0=1.2(0.5+W_0)-0.8$

$==>W_0=0.6+1.2W_0-0.8$

$==>0.2=0.2W_0$

$==>W_0=1eV$

hence,

The correct opt : B