$\begin {array} {1 1} (a)\;69 \times 10^{-26} J & \quad (b)\;4.31 \: neV \\ (c)\;3.78 \times 106{-28} J & \quad (d)\;2.36\: \mu\: eV \end {array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

Ans : (a)

For the $KE,\: K$ of an electron with velocity v, we have

$K=\large\frac{1}{2}\: m_ev^2$………..(i)

For de Broglie wavelength,

$ \lambda=h/m_ev$

$\Rightarrow \large\frac{v^2 = h^2 }{ \lambda^2m_e^2}$……….(ii)

Substituting eq. (ii) in (i),

$ K = \large\frac{m_eh^2}{2 \lambda^2m_e} \large\frac{h^2}{2\lambda^2m_e}$...........(iii)

$=\large\frac{ (6.6 \times 10^{-34} )^2 }{[2 \times (589 \times 10^{-9} )^2 \times 9.1 \times 10^{-31} ]}$

$≈ 69 \times 10^{-26}\: J$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...