(i) The expression for the energy difference between two electronic levels is given by

$$\overline v = {R_H}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right){Z^2}$$ ..... (ii)

$$\Delta E = hc\,.\,{R_H}\left( {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right){Z^2}$$ ..... (iii)

Z = 1 for H atom

Therefore, energy difference between first and second orbit of H - atom is given by

$$\Delta E = {R_H}\,.\,hc\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$

$$ = (1.09677 \times {10^7}\,{m^{ - 1}})(6.626 \times {10^{ - 34}}\,Js)(3 \times {10^8}\,m{s^{ - 1}}) \times \left( {{1 \over {{2^2}}} - {1 \over {{2^2}}}} \right)$$

$$ = 1.635 \times {10^{ - 18}}\,J$$

(ii) Energy of X-rays of wavelength

$$\lambda = {{hc} \over \lambda }$$

Energy emission for hydrogen like atom during transition from n = 2 to n = 1 energy state = $$1.635 \times {10^{ - 18}}\,{Z^2}J$$

$$\therefore$$ $$1.635 \times {10^{ - 18}}\,{Z^2}J = {{hc} \over \lambda }$$

$$ = {{(6.626 \times {{10}^{ - 34}}Js)(3 \times {{10}^8}\,m{s^{ - 1}})} \over {(3 \times {{10}^{ - 8}}\,m)}}$$

$$\therefore$$ Z = 2

(iii) The hydrogen atom-like species having atomic number 2 is He⁺