Question

The work functions of silver & sodium are $4.6\text{eV}$ and $2.3\text{eV}$ respectively. The ratio of the maximum kinetic energies of emitted electrons when illuminated by wavelength $155\text{nm}$ is $(hc=1240\text{eV nm})$

• A. $0.6$
• B. $0.3$
• C. $0.2$
• D. $0.95$

Answer 1

The correct option is A $0.6$

Given:
${\varphi }_{silver}=4.6\text{eV}$ ${\varphi }_{sodium}=2.3\text{eV}$

According to Einstein's photoelectric equation,

$K.E_{max}=E-\varphi$

$K.E_{max}=\frac{hc}{\lambda }-\varphi$

$\Rightarrow \frac{{\left(K{E}_{max}\right)}_{silver}}{{\left(K{E}_{max}\right)}_{sodium}}=\frac{\frac{hc}{\lambda }-{\varphi }_{silver}}{\frac{hc}{\lambda }-{\varphi }_{sodium}}$

$\Rightarrow \frac{{\left(K{E}_{max}\right)}_{silver}}{{\left(K{E}_{max}\right)}_{sodium}}=\frac{\frac{1240\text{eV nm}}{155\text{nm}}-4.6\text{eV}}{\frac{1240\text{eV nm}}{155\text{nm}}-2.3\text{eV}}$

$=\frac{8-4.6}{8-2.3}=\frac{3.4}{5.7}=0.596\simeq 0.6$

Hence, option $\left(A\right)$ is correct.

Why this question? Key Concept: According to Einstein's photoelectric equation, $K.E_{max}=E-\varphi =\frac{hc}{\lambda }-\varphi$