A Hydrogen-like atom has atomic number Z . Photons emitted in the...

From Einstein's photoelectric effect, we can write

E = K{E}_\max + ϕ ⇒ E = K{E}_\max + \dfrac{hc}{\lambda}

$⇒ ∆ {E}_{4to3} = 1 . 95eV + \dfrac{1240}{310}eV ⇒ ∆ {E}_{4to3} = 5 . 95eV$

Now, as we know for Hydrogen like atoms, we can write $∆ E = 13 . 6{Z}^{2}\left(\dfrac{1}{{\left({n}_{1}\right)}^{2}} - \dfrac{1}{{\left({n}_{2}\right)}^{2}}\right)$

Therefore,$⇒ 13 . 6{Z}^{2}\left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{4}^{2}}\right) = 5 . 95$

$⇒ 13 . 6{Z}^{2}\dfrac{7}{9 \times 16} = 5 . 95$

$⇒ {Z}^{2} = \dfrac{5 . 95 \times 9 \times 16}{13 . 6 \times 7} = 9$

$⇒ Z = 3$