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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
Detailed Solution
$h\nu = \phi_0 + K_{max}$
$h\nu = \phi_0 + 0.5$ ...(i)
$1.2h\nu = \phi_0 + 0.8$ ...(ii)
Solving, $\phi_0 = 1.0$ eV
$h\nu = \phi_0 + 0.5$ ...(i)
$1.2h\nu = \phi_0 + 0.8$ ...(ii)
Solving, $\phi_0 = 1.0$ eV