Question

Find the change in the angular momentum of the electron when it
jumps from the fourth orbit to the first orbit in a hydrogen atom​

Answer 1

Question :

Find the change in the angular momentum of the electron when it jumps from the fourth orbit to the first orbit in a hydrogen atom

Solution :

The electron jumps from four orbit to first orbit of the hydrogen atom

To finD

Change in angular momentum

According to Bohr's Postulates,

$\boxed{ \boxed{ \sf{L = n \dfrac{h}{2 \pi}}}}$

• n is the no.of orbits

For the fourth orbit,

$\large{ \leadsto \: \sf{L_{2} = 4 \dfrac{h}{2\pi}} }$

For the first orbit,

$\large{\leadsto \: \sf{ L_{1} = \dfrac{h}{2\pi} }}$

Change In Angular Momentum,

$\sf{ \Delta{L} = L_{2} - L_{1} } \\ \\ \longrightarrow \: \sf{ \Delta{L} = \frac{3h}{2\pi}} \\ \\ \longrightarrow \: \sf{ \Delta{L} = \dfrac{3 \times 6.62 \times {10}^{34} }{2 \times 3.14} } \\ \\ \longrightarrow \: \boxed{ \boxed{\sf{ \Delta{L} = 3.16 \times {10}^{ - 34} \: \: {Js}^{ - 1} }}}$

Answer 2

Answer:

$\large\boxed{\sf{ 3.17 \times {10}^{ - 34} \:\:J\:s}}$

Explanation:

It's being given that an electron of a hydrogen atom jumps from the fourth orbit to the first orbit.

To find the change in angular momentum:

Now, according to Bohr's second postulate,

Angular momentum (L) is given by,

$\large \boxed{ \sf{ \pink{L = \dfrac{nh}{2\pi}} }}$

where,

• n = Principal quantum number
• h = planks constant

So, we have,

For the fourth orbit, n = 4

$= > L_{4}= \dfrac{4h}{2\pi}$

Similarly, we have,

For the first orbit, n = 1

$=> L_{1} = \dfrac{h}{2 \pi}$

Now, Change in angular momentum,

$= > \triangle L= L_{4} - L_{1} \\ \\ = > \triangle L = \dfrac{h}{2\pi} (4 - 1) \\ \\ = > \triangle L = \dfrac{3 \times 6.64 \times {10}^{ - 34} }{2 \times 3.14} \\ \\ = > \sf{\triangle L = 3.17 \times {10}^{ - 34} \:\:J\:s}$