Question

A sample of hydrogen atoms gas contains 100 atoms All the atoms are excited to the seventh state. The total energy released by all the atoms is $\frac{4800}{49}$ Ry( Where Ry = 13 eV) then discuss the following things

a)Maximum energy of the emitted photon will be less then $\frac{4800}{49}Ry$

b)Maximum energy of the emitted photon may be grater than $\frac{48}{49}Ry$

c)The value of n = 6

d)total number of photon that can be emitted this sample may be less than 600

Answer 1

Given that,

Number of atoms, $n=100$

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

Maximum energy of one photon is

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

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

$n^{\prime }=7$

$\therefore n=6$

Maximum number of photons that can be emitted is 600.