Q.44. Angular momentum of an electron in a hydrogen atom is , then the energy of the electron is ___

Q: What is the correct answer for this exam Physics Q44?

The correct answer is C. Sol. According to Bohr's model, the angular momentum of an electron in a hydrogen atom is quantized as $L = \frac{nh}{2\pi}$. Given $L = \frac{3h}{\pi}$. So $\frac{nh}{2\pi} = \frac{3h}{\pi} \Rightarrow \frac{n}{2} = 3 \Rightarrow n = 6$. The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$. For $n=6$, $E_6 = -\frac{13.6}{6^2} = -\frac{1

Q: Q.44. Angular momentum of an electron in a hydrogen atom is , then the energy of the electron is _______ eV.

The correct answer is C. Sol. According to Bohr's model, the angular momentum of an electron in a hydrogen atom is quantized as $L = \frac{nh}{2\pi}$. Given $L = \frac{3h}{\pi}$. So $\frac{nh}{2\pi} = \frac{3h}{\pi} \Rightarrow \frac{n}{2} = 3 \Rightarrow n = 6$. The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$. For $n=6$, $E_6 = -\frac{13.6}{6^2} = -\frac{1

Q: What are the options for this this exam Physics question?

A. -1.51 · B. -0.85 · C. -0.38 · D. -0.28

Question

Q.44. Angular momentum of an electron in a hydrogen atom is , then the energy of the electron is _______ eV.

FirstInTest · Accepted Answer

Sol. According to Bohr's model, the angular momentum of an electron in a hydrogen atom is quantized as $L = \frac{nh}{2\pi}$. Given $L = \frac{3h}{\pi}$. So $\frac{nh}{2\pi} = \frac{3h}{\pi} \Rightarrow \frac{n}{2} = 3 \Rightarrow n = 6$. The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$. For $n=6$, $E_6 = -\frac{13.6}{6^2} = -\frac{13.6}{36} \approx -0.3777 	ext{ eV}$. Rounding to two decimal places, $E_6 \approx -0.38 	ext{ eV}$.

Answer

-1.51

Answer

-0.85

Answer

-0.38

Answer

-0.28

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