Photons of energy 1 eV and 2.5 eV successively illuminated a metal whose work function is 0.5 eV. The ratio of maximum speeds of emitted electron is

Question

Photons of energy 1 eV and 2.5 eV successively illuminated a metal whose work function is 0.5 eV. The ratio of maximum speeds of emitted electron is (A) 1:3 (B) 1:2 (C) 3:1 (D) 2:1

Tardigrade Admin · Accepted Answer

Energy of photon E=φ+(1/2)mv2 Vmax =√(2 (E - φ)/m) So putting values (Vmax 1/Vm a x 2 )=√(1 - 0.5/2.5 - 0.5) (Vmax 1/Vmax â¡ 2)=(1/2)

Q. Photons of energy 1 eV and 2.5 eV successively illuminated a metal whose work function is 0.5 eV. The ratio of maximum speeds of emitted electron is

$1 : 3$

$1 : 2$

$3 : 1$

$2 : 1$

Energy of photon $E = ϕ + \frac{1}{2} m v^{2}$
$V_{ma x} = \frac{2 ( E - ϕ )}{m}$
So putting values
$\frac{V _{ma x 1}}{V _{ma x 2}} = \frac{1 - 0.5}{2.5 - 0.5}$
$\frac{V _{ma x 1}}{V _{ma x 2}} = \frac{1}{2}$