Question

40. If electron falls from n = 3 to n = 2, then emitted energy is
(2). 12.09 eV (3) 1.9 eV
(1) 10.2 eV
(4) 0.65 eV​

Answer 1

$\bigstar$ Given

Electron falls from n = 3 to n = 2

$\bigstar$ To Find

The energy emitted

$\bigstar$ Solution

We know that

$\red{\sf \longrightarrow E=13.6 \left[\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right] \ eV}$

Here

• E = energy
• $\sf n_{f}$ = final energy level
• $\sf n_{i}$ = initial energy level

According to the question

We are asked to find the energy emitted

Therefore

We must find "E"

Given that

Electron falls from n = 3 to n = 2

Hence

• $\sf n_{f}$ = 2
• $\sf n_{i}$ = 3

Substituting the values

We get

$\green{\sf \longrightarrow E=13.6 \left[\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}} \right] \ eV}$

$\orange{\sf \longrightarrow E=13.6 \left[\dfrac{1}{4}-\dfrac{1}{9} \right] \ eV}$

On further simplification

We get

$\pink{\sf \longrightarrow E=1.88 \ eV}$

Hence

$\red{\sf \longrightarrow E \approx 1.9 \ eV}$

Therefore

$\checkmark$ Option (3) is correct