Discussion Forum

 A hydrogen atom in its ground state is irradiated by light of wavelength 970Å. Taking $hc/e =1.237\times 10^{-6} eVm$ and the ground state energy of hydrogen atom as - 13.6 eV the number of lines present in the emission spectrum is

Answer:

Explanation:

Energy of incident light (in eV) $E=\frac{12375}{970}=12.7 eV$

After excitation, let the electron jumps to nth State ,then $\frac{-13.6}{n^{2}}=-13.6+12.7$

Solving this equation, we get n=4;

Total number of lines in emission spectrum ,$ =  \frac{n(n-1)}{2}=\frac{4(4-1)}{2}=6$