Question

A photometal is illuminated by lights of wavelengths ${\lambda }_{1}and{\lambda }_2$ respectively. The maximum kinetic enegies of electrons emitted in the two cases are $E_1$ and $E_2$ respectively. The work function of metal is

• A. $\frac{E_2{\lambda }_1-E_1{\lambda }_2}{{\lambda }_1}$
• B. $\frac{E_1{\lambda }_1-E_2{\lambda }_2}{{\lambda }_1+{\lambda }_2}$
• C. $\frac{E_1{\lambda }_1+E_2{\lambda }_2}{{\lambda }_1-{\lambda }_2}$
• D. $\frac{E_2{\lambda }_2-E_1{\lambda }_1}{{\lambda }_1-{\lambda }_2}$

Answer 1

Answer: (D)

$\frac{E_2{\lambda }_2-E_1{\lambda }_1}{{\lambda }_1-{\lambda }_2}$

$E_1=\frac{hc}{{\lambda }_1}-\varphi ----------\left(1\right)$

$\Rightarrow (E_1+\varphi ){\lambda }_1=hc----------\left(2\right)$
$E_2=\frac{hc}{{\lambda }_2}-\varphi -------------\left(3\right)$
Now, substituting value of hc from (2) to (3)
$E_2=\frac{(E_1+\varphi ){\lambda }_1}{{\lambda }_2}-\varphi$
$E_2{\lambda }_2=E_1{\lambda }_1+\varphi {\lambda }_1-\varphi {\lambda }_2$
$\frac{E_2{\lambda }_2-E_1{\lambda }_1}{{\lambda }_1-{\lambda }_2}=\varphi$