A monochromatic radiation of wavelength lambd | vedclass

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(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $\frac{n(

Vedclass · Accepted Answer

$97.4$

Vedclass · Answer

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $\frac{n(n-1)}{2} = 6$.Solving for $n$: $n^2 - n - 12 = 0$, which factors to $(n-4)(n+3) = 0$. Since $n$ must be positive, $n = 4$.This means the hydrogen atoms are excited to the $n = 4$ energy level.The energy required to excite an electron from the ground state $(n_1 = 1)$ to the $n_2 = 4$ state is given by: $\Delta E = E_4 - E_1 = -13.6 \left( \frac{1}{4^2} - \frac{1}{1^2} ight) 	ext{ eV}$.$\Delta E = -13.6 \left( \frac{1}{16} - 1 ight) = -13.6 \left( -\frac{15}{16} ight) = 12.75 	ext{ eV}$.The wavelength $\lambda$ of the incident radiation is calculated using $\lambda = \frac{hc}{\Delta E}$.Given $hc = 1242 	ext{ eV-nm}$, we have $\lambda = \frac{1242}{12.75} \approx 97.4 	ext{ nm}$.

$A$ monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is. [Use $hc = 1242 \text{ eV-nm}$] (in $\text{ nm}$)

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