Solveeit Logo
Login

Question

Question: An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV. The ki...

An electron is in an excited state in a hydrogen like atom. It has a total energy of -3.4 eV. The kinetic energy of the electron is E and its de Broglie wavelength is Then

A

E=6.8 eV, λ=6.6×10−10 mE = 6.8\text{ eV, }\lambda = \text{6.6} \times 10^{- 10}\text{ m}

B

E=3.4 eV, λ=6.6×10−10 mE = 3.4\text{ eV, }\lambda = \text{6.6} \times 10^{- 10}\text{ m}

C

E=3.4 eV, λ=6.6×10−11 mE = 3.4\text{ eV, }\lambda = \text{6.6} \times 10^{- 11}\text{ m}

D

E=6.8 eV, λ=6.6×10−11 mE = 6.8\text{ eV, }\lambda = \text{6.6} \times 10^{- 11}\text{ m}

Answer

E=3.4 eV, λ=6.6×10−10 mE = 3.4\text{ eV, }\lambda = \text{6.6} \times 10^{- 10}\text{ m}

Explanation

Solution

The potential energy = -2× kinetic energy

=−2E= - 2E

∴\therefore Total energy =−2E+E=−E=−3.4eV= - 2E + E = - E = - 3.4eV

Or E=3.4eVE = 3.4eV

Let p = momentum and m = mass of the electron

∴E=p22mor\therefore E = \frac{p^{2}}{2m}or p =2mE= \sqrt{2mE}

De Broglie wavelength,

λ=hp=h2mE\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}

On substituting the values we get

λ=6.63×10−342×9.1×10−31×3.4×1.6×10−19\lambda = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 3.4 \times 1.6 \times 10^{- 19}}}

=6.6×10−10m= 6.6 \times 10^{- 10}m