Hydrogen atom is excited from ground state to another state with principal quantum number equal to $$4$$. Then the number of spectral lines in the emission spectra will be :
Explanation
Hint: Energy of photons = $$E=\frac{h c}{\lambda}$$
Solution:
Step 1: Find the energy of photon.
Given: $$h c=12500 \mathrm{eV}-A^{\circ}, \lambda=980 A^{\circ}$$
Energy of photons = $$E=\frac{h c}{\lambda}$$
$$\Rightarrow E=\frac{12500}{980}=12.75 \mathrm{eV}$$
Step 2: Find the excited state
Now, Energy of hydrogen atom in nth shell =$$E_{n}=\frac{-13.6}{n^{2}}$$
Energy absorbed by hydrogen atom = energy of photons
$$\Rightarrow E_{n}-E_{1}=E$$
$$\Rightarrow \frac{-13.6}{n^{2}}-\left[\frac{-13.6}{1^{2}}\right]=12.75 \mathrm{eV}$$
$$\Rightarrow 13.6-\frac{13.6}{n^{2}}=12.75$$
$$\Rightarrow 1-\frac{1}{n^{2}}=\frac{12.75}{13.6}$$
$$\Rightarrow 1-\frac{1}{n^{2}}=0.9375$$
$$\Rightarrow \frac{1}{n^{2}}=1-0.9375=0.0625$$
Step 3: Find radius in the excited state.
Radius of hydrogen atom = $$r=n^{2} a_{0}$$
$$\Rightarrow r=(4)^{2} a_{0}$$
$$\Rightarrow r=16 a_{0}$$
So, Radius of hydrogen atom in excited state is $$16 a_{0}$$
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