Monochromatic radiation of wavelength lambda | vedclass

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(A) 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

Vedclass · Accepted Answer

$95$

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(A) 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 incident photon corresponds to the transition from the ground state $(n=1)$ to the $n=5$ state.The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} ight)$.$\frac{1}{\lambda} = R \left( 1 - \frac{1}{25} ight) = R \left( \frac{24}{25} ight)$.Using $R \approx 1.097 	imes 10^7 \ m^{-1}$,we get $\lambda = \frac{25}{24 	imes 1.097 	imes 10^7} \approx 9.5 	imes 10^{-8} \ m = 95 \ nm$.

Monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample containing atoms in the ground state. Hydrogen atoms absorb the light and subsequently emit radiations of ten different wavelengths. The value of $\lambda$ is.....$nm$.

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