Question

A hydrogen atom initially in the ground level absorbs a photon and is excited n = 4 level then the wavelength of photon is

• A. 790 Å
• B. 870 Å
• C. 970 Å
• D. 1070 Å

Answer 1

Answer: (C)

970 Å

Answer 2

Here $n_{1} = 1$ and $n_{2} = 4$

Energy of photon absorbed, $E = E_{2} - E_{1}$

Since, $E_{n} = - \frac{13.6}{n^{2}}eV$

Then $E_{2} - E_{1} = - \frac{13.6}{(4)^{2}} - \left( - \frac{13.6}{(1)^{2}} \right)$

$= \frac{13.6}{16} + 13.6 = \frac{13.6 \times 15}{16}eV$

$= 12.75eV = 12.75 \times 1.6 \times 10^{- 19}J$

$= 20.4 \times 10^{- 18}J$

$E_{2} - E_{1} = \frac{hc}{\lambda}$

$\therefore\lambda = \frac{hc}{E_{2} - E_{1}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{20.4 \times 10^{- 18}}$

$= 9 - 70 \times 10^{- 8}m = 970 \times 10^{- 10} = 970Å$