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A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be
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The correct answer is:
$975 Å$
Number of possible spectral lines emitted when an electron jumps back to ground
$$
\text { state from } \mathrm{n}^{\text {th }} \text { orbit }=\frac{\mathrm{n}(\mathrm{n}-1)}{2}
$$
Here, $\frac{\mathrm{n}(\mathrm{n}-1)}{2}=6 \Rightarrow \mathrm{n}=4$
Wavelength $\lambda$ from transition from $\mathrm{n}=1$ to $\mathrm{n}=4$ is given by,
$$
\frac{1}{\lambda}=R\left(\frac{1}{1}-\frac{1}{4^{2}}\right) \Rightarrow \lambda=\frac{16}{15 R}=975 Å
$$
$$
\text { state from } \mathrm{n}^{\text {th }} \text { orbit }=\frac{\mathrm{n}(\mathrm{n}-1)}{2}
$$
Here, $\frac{\mathrm{n}(\mathrm{n}-1)}{2}=6 \Rightarrow \mathrm{n}=4$
Wavelength $\lambda$ from transition from $\mathrm{n}=1$ to $\mathrm{n}=4$ is given by,
$$
\frac{1}{\lambda}=R\left(\frac{1}{1}-\frac{1}{4^{2}}\right) \Rightarrow \lambda=\frac{16}{15 R}=975 Å
$$
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