$\begin {array} {1 1} (a)\;1.24\: BeV & \quad (b)\;12.4\: BeV \\ (c)\;1.98\: BeV & \quad (d)\;19.8\: BeV \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (a)

$ \lambda = 10^{-15}m$

Rest mass energy of electron: $m_oc^2 = 0.511\: MeV$

$= 0.511 \times 10^6 \times 1.6 \times 10^{-19}$

$= 0.8176 \times 10^{-13} J$

Momentum of a proton, $p = \large\frac{h}{\lambda}$

$= \large\frac{6.6 \times 10^{-34}}{ 10^{-15}} = 6.6 \times 10^{-19} kg m/s$

Relativistic relation for energy (E) is :

$E^2 = p^2c^2 + m^2_oc^4$

$= (6.6 \times 10^{-19} \times 3 \times 10^8 )^2 + (0.8176 \times 10^{-13} )^2$

$= 392.04 \times 10^{-22} + 0.6685 \times 10^{-26}$

$≈ 392.04 \times 10^{-22}$

So, $E = \large\frac{1.98 \times 10^{-10} J}{ 1.6 \times 10^{-19}} = 1.24 \times 10^9 \: eV = 1.24\: BeV$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...