Question

Atoms of a hydrogen like gas are in a particular excited energy level. When these atoms deexcite they emit photons of different energies. Maximum and minimum energies of emitted photons are E_max = 52.224 eV and E_min = 1.224 eV respectively. Identify the gas i.e. value of atomic number –

• A. 2
• B. 4
• C. 6
• D. 3

Answer 1

Answer: (A)

2

Answer 2

E_max = E_n – E₁ = 52.224

E_min = E_n – E_{(n – 1)} = 1.224 eV

energy of n^th level = E_n =$\frac{E_{1}}{n^{2}}$

$\frac{E_{1}}{n^{2}}$ – E₁ = 52.224 …(1)

$\frac{E_{1}}{n^{2}}$ – $\frac{E_{1}}{(n - 1)^{2}}$ = 1.224 …(2)

Solve for E₁ and n

E₁ = – 54.4 eV n = 5s