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Q.

Imagine that the electron in a hydrogen atom is replaced by a muon $(μ)$ . The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :

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a

331.2eV

b

2815.2 eV

c

27.2 eV

d

13.6eV

answer is B.

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Detailed Solution

$B o h r' s P o s t u l a t e m \frac{v^{2}}{r} = k \frac{e^{2}}{r^{2}} a n d m v r = n \frac{h}{2 π} e l i m i n a t i n g v, \Rightarrow r \propto \frac{1}{m} S i n c e, I o n i z a t i o n e n e r g y = E = \frac{1}{2} k \frac{e^{2}}{r} \Rightarrow E \propto \frac{1}{r} \propto m \Rightarrow \frac{E_{1}}{E_{2}} = \frac{m_{1}}{m_{2}} \Rightarrow \frac{13.6}{E_{2}} = \frac{m}{207 m} \Rightarrow E_{2} = 207 \times 13.6 = 2815.2 e V$

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