The energy of an electron in an orbit of the Bohr’s atom is where is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck’s constant,

Question

The energy of an electron in an orbit of the Bohr’s atom is $- 0.04 E_{0} e V$ where $E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck’s constant, then $\frac{2 π L}{h} =$ _____. [2026]

Smart Achievers · Accepted Answer

(1)

$Angular momentum L = \frac{n h}{2 π}$

$n = \frac{2 π L}{h}$

$Energy E = - \frac{13.6}{n^{2}} Z^{2}$

$E \Rightarrow - \frac{E_{0}}{n^{2}} = - 0.04 E_{0}$

$n^{2} = 25, n = 5$

Solution

Q.
The energy of an electron in an orbit of the Bohr’s atom is $- 0.04 E_{0} e V$ where $E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck’s constant, then $\frac{2 π L}{h} =$ _____. [2026]

1 5

2 2

3 6

4 4

Ans.
(1)

$Angular momentum L = \frac{n h}{2 π}$

$n = \frac{2 π L}{h}$

$Energy E = - \frac{13.6}{n^{2}} Z^{2}$

$E \Rightarrow - \frac{E_{0}}{n^{2}} = - 0.04 E_{0}$

$n^{2} = 25, n = 5$

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Solution

Q. The energy of an electron in an orbit of the Bohr’s atom is -0.04E0 eV where E0 is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck’s constant, then 2πLh=_____. [2026]

1 5

2 2

3 6

4 4

Ans. (1) Angular momentum L=nh2π n=2πLh Energy E=-13.6n2Z2 E⇒-E0n2=-0.04E0 n2=25, n=5

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Q.
The energy of an electron in an orbit of the Bohr’s atom is $- 0.04 E_{0} e V$ where $E_{0}$ is the ground state energy. If L is the angular momentum of the electron in this orbit and h is the Planck’s constant, then $\frac{2 π L}{h} =$ _____. [2026]

Ans.
(1)

$Angular momentum L = \frac{n h}{2 π}$

$n = \frac{2 π L}{h}$

$Energy E = - \frac{13.6}{n^{2}} Z^{2}$

$E \Rightarrow - \frac{E_{0}}{n^{2}} = - 0.04 E_{0}$

$n^{2} = 25, n = 5$