A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take hc =1230 eV - nm ).

Question

A metal exposed to light of wavelength 800 nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500 nm is used. The work function of the metal is (Take hc =1230 eV - nm ). (A) 1.537 eV (B) 2.46 eV (C) 0.615 eV (D) 1.23 eV

Tardigrade Admin · Accepted Answer

k 1=(1230/800)-φ..... (1) k 2=2 k 1=(1230/500)-φ.....(2) Eliminating k 1 from (1) and (2) we get 0=(1230/500)-(1230/400)+φ φ=0.615 eV

Q. A metal exposed to light of wavelength $800 nm$ and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength $500 nm$ is used. The work function of the metal is (Take $h c = 1230 e V - nm)$ .

$1.537 e V$

$2.46 e V$

$0.615 e V$

$1.23 e V$

$k_{1} = \frac{1230}{800} - ϕ .....$ (1)
$k_{2} = 2 k_{1} = \frac{1230}{500} - ϕ .....$ (2)
Eliminating $k_{1}$ from (1) and (2) we get
$0 = \frac{1230}{500} - \frac{1230}{400} + ϕ$
$ϕ = 0.615 e V$

Q. A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is (Take hc=1230eV−nm).

1.537eV

2.46eV

0.615eV

1.23eV