Question

A sample of hydrogen gas contain 100 atoms. All the atoms are excited to the same n excited state. The energy released by all the atoms is $\frac{4800}{49}$ . R = 13.6 eV), as they come to the ground state through various types of transition Then Find

• A. maximum energy of the emitted photon will be less than $\frac{48}{49}Rch.$
• B. Maximum energy of the emitted photon may be greater than $\frac{48}{49}Rch$
• C. The value of n = 6.
• D. None of these

Answer 1

The correct option is C The value of n = 6.

Given that,

Number of atoms, $n=100$

Energy released by all the atoms is $E=\frac{4800}{49}rch$

Maximum energy of one photon is

$\frac{\frac{4800}{49}}{100}=\frac{48}{49}$

$\frac{48}{49}rch=rch(\frac{1}{1^2}-\frac{1}{n{{}^{\prime }}^2})$

$n^{\prime }=7$

$\therefore n=6$

Maximum number of photons that can be emitted is 600.