An energy of 68.0 eV is required to excite a hydrogen-like atom in its second Bohr energy level to third energy level the charge of nucleus is Ze. The wavelength of a radiation required to eject the electron from first orbit to infinity is

2.2 nm

2.85 nm

3.2 nm

2.5 nm

Solution

2.5 nm

Given:-

n₁ =2,

n₂ = 3 and

ΔE = 68 eV

The difference in energies of two orbits

ΔE = 13.6 Z² $[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

ΔE = 13.6 Z² $[\frac{1}{2^{2}} - \frac{1}{3^{2}}]$

ΔE = 13.6 Z² $[\frac{1}{4} - \frac{1}{9}]$

ΔE = 12.6 Z² $[\frac{9 - 4}{36}]$

ΔE = 13.6 Z² $\times \frac{5}{36}$

Z² = $\frac{68 \times 36}{13.6 \times 5}$

Z = 6

$∵$ Wavelength of photon

$\frac{1}{λ}$ = Z²R $[\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$

n₁ = 1 and n₂ = ∞

$\frac{1}{λ}$ = 6² × R $[\frac{1}{1^{2}} - \frac{1}{\infty}]$

= 36R

λ = $\frac{1}{36 R}$

= $\frac{1}{36 \times 1.097 \times 10^{7}}$

= $\frac{10^{- 7}}{39.5} m$

λ = 0.025 × 10^-7 m

λ = 2.5 nm

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