Consider a hydrogen like atom whose energy in nth excited state is given by En=-(13.6Z2/n2), when this excited atom makes a transition from excited state to ground state, most energetic photons have energy E max =52.224 eV and least energetic photons have energy E min =1.224 eV. The atomic number of atom is

Question

Consider a hydrogen like atom whose energy in nth excited state is given by En=-(13.6Z2/n2), when this excited atom makes a transition from excited state to ground state, most energetic photons have energy E max =52.224 eV and least energetic photons have energy E min =1.224 eV. The atomic number of atom is (A) 2 (B) 4 (C) 5 (D) None of the above

Tardigrade Admin · Accepted Answer

Maximum energy is liberated for transition

E_{n} \to 1

and minimum energy for

E_{n} \to E_{n - 1}

Hence,

\frac{E _{1}}{n ^{2}} - E_{1} = 52.224 e V

..(i)
and

\frac{E _{1}}{n ^{2}} - \frac{E}{( n - 1 ) ^{2}} = 1.224 e V

..(ii)
Solving Eqs. (i) and (ii),
we get

E_{1} = - 54.4 e V

and

n = 5

But

E_{1} = - \frac{13.6 Z ^{2}}{1 ^{2}}

∴

- 54.4 = \frac{13.6}{1 ^{2}} Z^{2}

\Rightarrow

Z = 2

2

4

5

None of the above

Maximum energy is liberated for transition En​→1 and minimum energy for En​→En−1​ Hence, n2E1​​−E1​=52.224eV ..(i) and n2E1​​−(n−1)2E​=1.224eV ..(ii) Solving Eqs. (i) and (ii), we get E1​=−54.4eV and n=5 But E1​=−1213.6Z2​ ∴ −54.4=1213.6​Z2 ⇒ Z=2