The kinetic energy of an... | JEE Main 2023

JEE MAIN 2023

06-04-23 S1

Question

The kinetic energy of an electron, $\alpha$-particle and a proton are given as 4K, 2K and K respectively. The de-Broglie wavelength associated with electron ( $\lambda \mathrm{e}$ ) $\alpha$-particle ( $\lambda \alpha$ ) and the proton ( $\lambda \mathrm{p}$ ) are as follows .

Select the correct option:

$\lambda \alpha=\lambda \mathrm{p}<\lambda \mathrm{e}$

$\lambda \alpha>\lambda \mathrm{p}>\lambda \mathrm{e}$

$\lambda \alpha<\lambda \mathrm{p}<\lambda \mathrm{e}$

$\lambda \alpha=\lambda \mathrm{p}>\lambda \mathrm{e}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution