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(A) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For the ground state $(n=1)$,$E_1 = -1

Vedclass · Accepted Answer

$2.55 	imes 10^{15} \ Hz$

Vedclass · Answer

(A) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.For the ground state $(n=1)$,$E_1 = -13.6 \ eV$.For the second orbit $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.The energy of the emitted photon during the transition from $n=2$ to $n=1$ is $\Delta E = E_2 - E_1 = -3.4 \ eV - (-13.6 \ eV) = 10.2 \ eV$.The frequency $
u$ is given by $\Delta E = h
u$,so $
u = \frac{\Delta E}{h}$.Substituting the given values: $
u = \frac{10.2 \ eV}{4 	imes 10^{-15} \ eV \cdot s} = 2.55 	imes 10^{15} \ Hz$.

The frequency of light emitted,when the electron makes a transition from the level of principal quantum number $n=2$ to the level with $n=1$ is (Take,the ionization energy of hydrogen to be $13.6 \ eV$ and $h \simeq 4 \times 10^{-15} \ eV \cdot s$)

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