Electromagnetic radiation falls on a metallic body whose work function is 2 eV. For a particular radiation of frequency v, the maximum kinetic energy of the photoelectron is found to be 4 eV. What would be the maximum kinetic energy of photoelectron for the radiation of frequency (5 v/3) ?

Question

Electromagnetic radiation falls on a metallic body whose work function is 2 eV. For a particular radiation of frequency v, the maximum kinetic energy of the photoelectron is found to be 4 eV. What would be the maximum kinetic energy of photoelectron for the radiation of frequency (5 v/3) ? (A) (8/3) eV (B) 8 eV (C) (10/3) eV (D) (20/3) eV

Tardigrade Admin · Accepted Answer

h v=φ+K max h v=2+4=6 eV Now, h((5/3) v)=2+K max K max =(5/3) × 6-2 ⇒ K max =8 eV

$\frac{8}{3} e V$

$8 e V$

$\frac{10}{3} e V$

$\frac{20}{3} e V$

$h v = ϕ + K_{m a x}$ $h v = 2 + 4 = 6 e V$
Now, $h (\frac{5}{3} v) = 2 + K_{m a x}$
$K_{m a x} = \frac{5}{3} \times 6 - 2 \Rightarrow K_{m a x} = 8 e V$

Q. Electromagnetic radiation falls on a metallic body whose work function is 2eV. For a particular radiation of frequency v, the maximum kinetic energy of the photoelectron is found to be 4eV. What would be the maximum kinetic energy of photoelectron for the radiation of frequency 35v​ ?

38​eV

8eV

310​eV

320​eV

hv=ϕ+Kmax​ hv=2+4=6eV Now, h(35​v)=2+Kmax​ Kmax​=35​×6−2⇒Kmax​=8eV

$\frac{8}{3} e V$

$8 e V$

$\frac{10}{3} e V$

$\frac{20}{3} e V$

$h v = ϕ + K_{m a x}$ $h v = 2 + 4 = 6 e V$
Now, $h (\frac{5}{3} v) = 2 + K_{m a x}$
$K_{m a x} = \frac{5}{3} \times 6 - 2 \Rightarrow K_{m a x} = 8 e V$