Question

Photons of energy $1$eV and $2.5$eV successively illuminate a metal whose work function if $0.5$ eV. The ratio of maximum speed of emitted electrons is ________.

• A. $1:2$
• B. $2:1$
• C. $3:1$
• D. $1:3$

Answer 1

The correct option is A $1:2$

$Energy=W+K.E.$

where $W=$Work function

$K.E.=$ Kinetic energy of photon $=\frac{1}{2}m{v}^2$

$m=$mass of emitted electrons

$v=$velocity of emitted electrons

For proton 1:

Given:

$Energy=1eV$

$W=0.5eV$

$Energy=W+K.E.$

$1=0.5+K.E.$

$0.5=K.E$

$K.E.=\frac{1}{2}m{v}^2$

$0.5=\frac{1}{2}m{v}_1^2$ (1)

For proton 2:

Given:

$Energy=2.5eV$

$W=0.5eV$

$Energy=W+K.E.$

$2.5=0.5+K.E.$

$2=K.E$

$K.E.=\frac{1}{2}m{v}^2$

$2=\frac{1}{2}m{v}_2^2$ (2)

Dividing (1) by (2) gives

$\frac{0.5}{2}=\frac{\frac{1}{2}m{v}_1^2}{\frac{1}{2}m{v}_2^2}$

$\frac{1}{4}=\frac{v_1^2}{v_2^2}$

$\frac{1}{2}=\frac{v_1}{v_2}$

Hence the correct option is (A).