Q1)Calculate wavelength of the radiation emitted producing a line in the Lyman series when electron falls from 4th stationary state in hydrogen atom.

Q2)Calculate energy and radius of 2nd orbit of He+ion

Q3)The electron energy of hydrogen atom in the ground state is -2.18*10^-18 J per atom.Calculate what change will occur in the location of the electron if energy of 1.938*10^-18 per atom is supplied to the hydrogen atom.

Q4)Calculate the momentum of particle which has de Broglie wavelength of 1A.

1) For lyman series of spectrum transition from 4th state , n₁ = 1 , n₂ = 4 , putting it in the formula ,we get
$$\begin{gathered} \overline{\nu } =R\left(\frac{1}{1^2}-\frac{1}{4^2}\right) , R = 109677 c{m}^{-1} \\ \overline{\nu } =102822.18 c{m}^{-1} \\ \lambda = \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\overline{\nu }$}\right. = 9.725 \times {10}^{-6} cm = 97.25 nm \end{gathered}$$
2) 
For the radius and energy of a stationary state of Hydrogen atom ,the formula is , 
$$\begin{gathered} E_n = -\frac{1312}{n^2} kJ mo{l}^{-1} \\ {r}_n = a_o{n}^2 \end{gathered}$$ 
n = n^th​ stationary state , a_o = 52.9 pm and it is the radii of the first stationary state.
For n =2 
$$\begin{gathered} E_n = -\frac{1312}{n^2} kJ mo{l}^{-1} \\ {E}_n = -\frac{1312}{2^2} kJ mo{l}^{-1} \\ {E}_n = -328 kJ mo{l}^{-1} \\ And, \\ {r}_n = a_o{n}^2 = 52.9\times 2^2 = 211.6 pm \end{gathered}$$

3) Since , $1.938\times {10}^{-18} J per atom$ is given , change in energy from the ground state and the excited state will be $1.938\times {10}^{-18} J per atom$.

$$\begin{gathered} E = \frac{hc}{\lambda } \\ p = \frac{h}{\lambda } \\ E = pc ; \Delta E = \Delta p \times c \end{gathered}$$

$$\begin{gathered} \Delta E = 1.938 \times {10}^{-18} = \Delta p \times 3\times {10}^{8 } \\ \Delta p = \frac{1.938 \times {10}^{-18} }{3\times {10}^{8 }} = 0.646\times {10}^{-26} kg m se{c}^{-1} \\ According to uncertainity formula , \\ \Delta x \times \Delta p = \frac{h}{4\pi } \\ \Delta x = \frac{h}{4\pi \Delta p} = \frac{6.626\times {10}^{-34}}{4\times 3.14\times 0.646\times {10}^{-26}} = 0.816 \times {10}^{-8} m \approx {10}^{-8} m \end{gathered}$$
Approximately change of 10^-8 m in location will occur.

4) de Broglie wavelength  $\lambda = \frac{h}{p}$ 
$$\begin{gathered} \lambda = 1 \stackrel{o}{A} = {10}^{-10} m \\ h = 6.626\times {10}^{-34} kg m^2 se{c}^{-1} \\ p = \frac{h}{\lambda } \\ p = \frac{6.626\times {10}^{-34}}{{10}^{-10}} = 6.626 \times {10}^{-24} kg m se{c}^{-1} \end{gathered}$$