$\begin {array} {1 1} (a)\;E_X < E_R < E_M & \quad (b)\;E_X > E_M > E_R \\ (c)\;E_X < E_R > E_M & \quad (d)\;E_X > E_R > E_M \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$ \lambda$ of X-rays ranges from $1 \: A^{\circ}\: to\: 100\: A^{\circ},\lambda$ of microwaves ranges from $10^7A^{\circ}\: to 10^9A^{\circ} $ and $ \lambda$ of

radio waves ranges from $ 10^9A^{\circ}\: to\: 10^{14}A^{\circ}$

So, $\lambda_X < \lambda_M < \lambda_R$

Lesser the wavelength, more is the energy

So,$ E_X > E_M > E_R$

Ans : (b)

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...