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  • The energy of an excited hydrogen atom is . The angular momentum of the electron is: <img alt="\mathrm{(Take \le

The energy of an excited hydrogen atom is -3.4 \mathrm{eV}. The angular momentum of the electron is:

\mathrm{(Take \left.\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\right)}

Option: 1

1.05 \times 10^{-34} \mathrm{Js}


Option: 2

2.11 \times 10^{-34} \mathrm{Js}


Option: 3

3.16 \times 10^{-34} \mathrm{Js}


Option: 4

4.22 \times 10^{-34} \mathrm{Js}


Answers (1)

best_answer

The energy of nth state of hydrogen atom is

\begin{aligned} & \mathrm{E}_{\mathrm{n}}=-\frac{13.6}{\mathrm{n}^2} \mathrm{eV} \\ \\& \therefore \quad \mathrm{n}^2=\frac{-13.6 \mathrm{eV}}{\mathrm{E}_{\mathrm{n}}}=\frac{-13.6 \mathrm{eV}}{-3.4 \mathrm{eV}}=4\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{\left(\because \mathrm{E}_{\mathrm{n}}=-3.4 \mathrm{eV}\right)} \\ \\& \therefore \quad \mathrm{n}=2 \end{aligned}

Accordig to Bohr’s quantisation conditon

Angular momentum, \mathrm{\mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{2 \mathrm{~h}}{2 \pi}=\frac{\mathrm{h}}{\pi}=\frac{6.63 \times 10^{-34} \mathrm{Js}}{3.14}=2.11 \times 10^{-34} \mathrm{Js}}

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