$\begin{array}{1 1}(a)\;4.71A^{\large\circ}&(b)\;3.72A^{\large\circ}\\(c)\;4A^{\large\circ}&(d)\;5.6A^{\large\circ}\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

If $E_1=13.6eV$

$E_2=13.6\times 1.50eV=20.4eV$

$\Delta E=(E_2-E_1)=20.4-13.6=6.8eV$

$\Rightarrow 6.8\times 1.6\times 10^{-19}J$

$\lambda=\large\frac{h}{\sqrt{2m(KE)}}$

$\Rightarrow \large\frac{6.626\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 6.8\times 1.6\times 10^{-19}}}$

$\Rightarrow 4.71\times 10^{-10}m$

$\Rightarrow 4.71A^{\large\circ}$

Hence (a) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...