Question

In an excited state of hydrogen like atom an electron has a total energy of -3.4 eV. Then its de Broglie wavelength $\lambda$ is nearly

• A. $6.6\times {10}^{-10}m$
• B. $6.6\times {10}^{-11}m$
• C. $7.6\times {10}^{-11}m$
• D. $7.6\times {10}^{-10}m$

Answer 1

The correct option is A $6.6\times {10}^{-10}m$

$\lambda =\frac{h}{\sqrt{2mE}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9\times {10}^{-31}\times 3.4\times 1.6\times {10}^{-19}}}$
$=\frac{6.6\times {10}^{-34}}{12\times {10}^{-25}\times \sqrt{0.68}}=6.6\times {10}^{-10}m$