An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr quantization is applicable, then the radius of the electronic orbit in the first excited state is: [2025]

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Q.
An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr quantization is applicable, then the radius of the electronic orbit in the first excited state is: [2025]

1 $\sqrt{\frac{2 h}{π e B}}$

2 $\sqrt{\frac{4 h}{π e B}}$

3 $\sqrt{\frac{h}{2 π e B}}$

4 $\sqrt{\frac{h}{π e B}}$

Ans.
(4)

$m v r = \frac{n h}{2 π}$ ...(i)

$r = \frac{v m}{B q}$ ...(ii)

n = 2

$m r (\frac{r B q}{m}) = \frac{2 h}{2 π} \Rightarrow r = \sqrt{\frac{h}{π B q}}$

$q = e \Rightarrow r = \sqrt{\frac{h}{π B e}}$

Solution

Q.
An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr quantization is applicable, then the radius of the electronic orbit in the first excited state is: [2025]

1 $\sqrt{\frac{2 h}{π e B}}$

2 $\sqrt{\frac{4 h}{π e B}}$

3 $\sqrt{\frac{h}{2 π e B}}$

4 $\sqrt{\frac{h}{π e B}}$

Ans.
(4)

$m v r = \frac{n h}{2 π}$ ...(i)

$r = \frac{v m}{B q}$ ...(ii)

n = 2

$m r (\frac{r B q}{m}) = \frac{2 h}{2 π} \Rightarrow r = \sqrt{\frac{h}{π B q}}$

$q = e \Rightarrow r = \sqrt{\frac{h}{π B e}}$

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Solution

Q. An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr quantization is applicable, then the radius of the electronic orbit in the first excited state is: [2025]

1 2hπeB

2 4hπeB

3 h2πeB

4 hπeB

Ans. (4) mvr=nh2π ...(i) r=vmBq ...(ii) n = 2 mr(rBqm)=2h2π ⇒ r=hπBq q=e ⇒ r=hπBe

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Q.
An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr quantization is applicable, then the radius of the electronic orbit in the first excited state is: [2025]

1 $\sqrt{\frac{2 h}{π e B}}$

2 $\sqrt{\frac{4 h}{π e B}}$

3 $\sqrt{\frac{h}{2 π e B}}$

4 $\sqrt{\frac{h}{π e B}}$

Ans.
(4)

$m v r = \frac{n h}{2 π}$ ...(i)

$r = \frac{v m}{B q}$ ...(ii)

n = 2

$m r (\frac{r B q}{m}) = \frac{2 h}{2 π} \Rightarrow r = \sqrt{\frac{h}{π B q}}$

$q = e \Rightarrow r = \sqrt{\frac{h}{π B e}}$