Question

Mercury violet light $(\lambda =4558\stackrel{\circ }{A})$ is falling on a photosensitive material $(\varphi =2.5eV)$. The speed of the ejected electrons is in $m{s}^{-1}$, about

• A. $3\times {10}^5$
• B. $2.65\times {10}^5$
• C. $4\times {10}^4$
• D. $3.65\times {10}^7$

Answer 1

The correct option is B

$2.65\times {10}^5$

By using $E=W_0+\frac{1}{2}m{v}_{max}^2;whereE=\frac{12375}{4558}=2.71eV$
$\Rightarrow 2.71eV=2.5eV+\frac{1}{2}\times 9.1\times {10}^{-31}\times v_{max}^2$
$\Rightarrow 0.21\times 1.6\times {10}^{-19}=\frac{1}{2}\times 9.1\times {10}^{-31}\times v_{max}^2$
$\Rightarrow v_{max}=2.65\times {10}^{5}m/s$