Monochromatic radiation is incident on a hydr | vedclass

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(D) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $N = \fra

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$95.1$

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(D) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $N = \frac{n(n-1)}{2}$.Given $N = 10$, we have $\frac{n(n-1)}{2} = 10$, which implies $n^2 - n - 20 = 0$.Solving the quadratic equation, $(n-5)(n+4) = 0$, we get $n = 5$ (since $n$ must be positive).The energy of the $n=5$ state is $E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} = -0.544 	ext{ eV}$.The energy of the ground state $(n=1)$ is $E_1 = -13.6 	ext{ eV}$.The energy of the incident photon required to excite the atom from $n=1$ to $n=5$ is $\Delta E = E_5 - E_1 = -0.544 - (-13.6) = 13.056 	ext{ eV}$.The wavelength $\lambda$ is given by $\lambda = \frac{hc}{\Delta E} = \frac{1242 	ext{ eV-nm}}{13.056 	ext{ eV}} \approx 95.1 	ext{ nm}$.

Monochromatic radiation is incident on a hydrogen $(H)$ sample which is in the ground state. If the hydrogen atoms emit radiation of $10$ different wavelengths after absorbing the incident radiation, then the wavelength of the incident radiation is (Let $hc = 1242 \text{ eV-nm}$) (in $\text{ nm}$)

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