If the wavelength for an electron emitted from H-atom is $3.3\times {10}^{-10}\mathrm{m}$, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is _______ times. 

(Nearest integer)

[Given : h=6.626 X 10^-34 Js,

         Mass of electron = 9.1 X 10^-31]

1. Calculate the Energy of the Emitted Electron:

The energy of the emitted electron is given by the formula:

E = h² / (2mλ²)

where:

• h = 6.626 × 10⁻³⁴ Js (Planck's constant)
• m = 9.1 × 10⁻³¹ kg (mass of the electron)
• λ = 3.3 × 10⁻¹⁰ m (wavelength of the electron)

Substitute the values:

E = (6.626 × 10⁻³⁴)² / [2 × 9.1 × 10⁻³¹ × (3.3 × 10⁻¹⁰)²]

First, calculate the numerator:

h² = (6.626 × 10⁻³⁴)² = 4.39 × 10⁻⁶⁷

Next, calculate the denominator:

2mλ² = 2 × 9.1 × 10⁻³¹ × (3.3 × 10⁻¹⁰)² = 1.98 × 10⁻⁴⁹

Now, divide:

E = 4.39 × 10⁻⁶⁷ / 1.98 × 10⁻⁴⁹ = 2.22 × 10⁻¹⁸ J

2. Find the Minimum Energy Required for Escape (Ionization Energy):

The ionization energy of hydrogen in its ground state is:

E_ionization = 13.6 eV

Convert this to joules:

E_ionization = 13.6 × 1.602 × 10⁻¹⁹ = 2.18 × 10⁻¹⁸ J

3. Find the Ratio of Energy Absorbed:

The total energy absorbed by the electron is the sum of the ionization energy and the kinetic energy (E) of the emitted electron. The ratio of the total absorbed energy to the ionization energy is:

Ratio = (E_ionization + E) / E_ionization

Substitute the values:

Ratio = (2.18 × 10⁻¹⁸ + 2.22 × 10⁻¹⁸) / 2.18 × 10⁻¹⁸

Simplify:

Ratio = 4.4 × 10⁻¹⁸ / 2.18 × 10⁻¹⁸ ≈ 2.02

Final Answer:

The nearest integer is: 2