Question

Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda =975\stackrel{\circ }{A}$. Number of spectral lines in the resulting spectrum emitted will be

• A. 3
• B. 2
• C. 6
• D. 10

Answer 1

Answer: (C)

Step 1: Given

Wavelength , $\lambda =975A^o$

Step 2: Formula used and Calculation

Energy incident $=\frac{hc}{\lambda }$
Energy incident $=\frac{hc}{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-19}}eV$
$=12.75eV$

Now,

Energy for 1st orbit is $-13.6eV$

Energy for 4th orbit is $-0.85eV$

Difference =$12.75eV$
The hydrogen atom will be excited to n = 4
Therefore, number of spectral lines $=\frac{4(4-1)}{2}=6$