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# What is the wavelength of light for the least energetic photon emitted photon in the Lyman series of the hydrogen spectrum. (take $hc = 1240{\text{ eVnm}}$)(A) $82{\text{ nm}}$(B) $102{\text{ nm}}$(C) $122{\text{ nm}}$(D) $150{\text{ nm}}$

Last updated date: 15th Sep 2024
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Hint To solve this question, we need to use the Rydberg formula for the frequency of radiations emitted for the Lyman series of the hydrogen atom. Then, we have to apply the required condition on that formula to get the required value of wavelength.
The formulae which are used to solve this question are given by
$\Rightarrow \nu = Rc\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{n^2}}}} \right)$, here $\nu$ is the frequency of the radiation emitted corresponding to the Lyman series of the hydrogen spectrum, $R$ is the Rydberg’s constant, $c$ is the speed of light, and $n = 2,3,4,...$
$\Rightarrow E = h\nu$, here $E$ is the energy corresponding to the radiation having frequency $\nu$ and, $h$ is the Planck’s constant.

We know from the Planck’s equation that the energy of photon emitted is related to its frequency by
$\Rightarrow E = h\nu$ (1)
According to the question, we have the least energetic photon being emitted. From (1) the energy is proportional to the frequency. So, if the energy is minimum, then this means that frequency of the photon is minimum.
Now, for the Lyman series of the hydrogen atom, we have the formula
$\Rightarrow \nu = Rc\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{n^2}}}} \right)$, $n = 2,3,4,...$
As the frequency is minimum so we take $n = 2$. So, we have
$\Rightarrow \nu = Rc\left( {\dfrac{1}{{{1^2}}} - \dfrac{1}{{{2^2}}}} \right)$
$\Rightarrow \nu = \dfrac{3}{4}Rc$ (2)
Now, as we know that the wavelength of the radiation is related with its frequency by
$\Rightarrow \lambda \nu = c$
$\Rightarrow \lambda = \dfrac{c}{\nu }$ (3)
Substituting (2) in (3) we get
$\Rightarrow \lambda = \dfrac{4}{{3Rc}}c$
$\Rightarrow \lambda = \dfrac{4}{{3R}}$
We know that $R = 1.09 \times {10^7}{m^{ - 1}}$. Substituting it above, we get
$\Rightarrow \lambda = \dfrac{4}{{3 \times 1.09 \times {{10}^7}}}m$
On solving we get
$\Rightarrow \lambda = 1.22 \times {10^{ - 7}}m$
$\Rightarrow \lambda = 122 \times {10^{ - 9}}m$
We know that $1nm = {10^{ - 9}}m$. Therefore
$\Rightarrow \lambda = 122nm$
Thus the wavelength of light for the least energetic photon emitted photon in the Lyman series of the hydrogen spectrum is equal to $122nm$.
Hence, the correct answer is option C.

Note
We could also use the formula for the energy in the nth state for the hydrogen atom by applying the condition for the Lyman series. Proceeding in this way, we could use the value $hc = 1240{\text{ eVnm}}$which is given in the question.