$\begin{array}{1 1}(a)\;6.54\times 10^5m/s&(b)\;5.5\times 10^5m/s\\(c)\;6.54\times 10^2m/s&(d)\;4.63\times 10^3m/s\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Using Einstein's photoelectric equation, $E_1=KE+W$

$E_1=h$$\large\frac{c}{\lambda}=\frac{(6.63\times 10^{-34})\times(3\times 10^8)}{200\times 10^{-9}}$

Given, $W=5eV=5\times (1.6\times 10^{-19})J = 8.0\times 10^{-19}J$

$KE=E_1-W= 9.945\times 10^{-19}-8.0\times 10^{-19} = 1.945\times 10^{-19}J$

Now, Kinetic energy of the electron, $KE=\large\frac{1}{2}$$mv^2$

$\Rightarrow v=\sqrt{\large\frac{2KE}{m}}$

$\Rightarrow v=\sqrt{\large\frac{2(1.195\times 10^{-19})}{9.1\times 10^{-31}}}$

$\Rightarrow 6.54\times 10^2m/s$

Hence (c) is the correct answer.

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...