Question

A free hydrogen atom after absorbing a photon of wavelength ${{\lambda }_{a}}$ gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength ${{\lambda }_{e}}$. Let the change in momentum of the atom due to the absorption and the emission are $\Delta {{p}_{a}},\Delta {{p}_{e}}$, respectively. If $\dfrac{{{\lambda }_{a}}}{{{\lambda }_{e}}}=\dfrac{1}{5}$. Which of the option(s) is/are correct ?
A) $\begin{aligned} & {{\lambda }_{e}}=418nm \\\ \end{aligned}$
B) ratio of kinetic energies of electrons in state n=m to n=1 is one fourth.
C) m=2
D) $\dfrac{\Delta {{P}_{a}}}{\Delta {{P}_{e}}}=\dfrac{1}{2}$

Answer 1

As the ratio of wavelengths of the two different photons are given, we can find the total energy of the photon in terms of wavelength and find the value of m. Next, we can find the energy emitted when the electron jumped from n=4 to n=m state. Finally, we can find the ratio of momentums easily.

Complete step by step answer:
Let us write down the given terms,
$\dfrac{{{\lambda }_{a}}}{{{\lambda }_{e}}}=\dfrac{1}{5}$
Energy of the photon that excited from n=1 to n=4,
$E=\dfrac{hc}{{{\lambda }_{a}}}=13.6(1-\dfrac{1}{16})....(1)$
Energy of the photon that jumped from n=4 to n=m,
$E=\dfrac{hc}{{{\lambda }_{e}}}=13.6(\dfrac{1}{{{m}^{2}}}-\dfrac{1}{16})...(2)$
Dividing equations one and two, we get,
$\begin{aligned} & \dfrac{1}{5}=\dfrac{1}{{{m}^{2}}}-\dfrac{1}{16}\times \dfrac{16}{15} \\\ & \Rightarrow \dfrac{1}{{{m}^{2}}}=\dfrac{4}{16} \\\ & \therefore m=2 \\\ \end{aligned}$
Now, the energy of the photon and its wavelength is,
$\begin{aligned} & {{\lambda }_{e}}=\dfrac{12400\times 16}{13.6\times 3} \\\ & \Rightarrow {{\lambda }_{e}}=4862 \\\ \end{aligned}$
Now, we have,
$\begin{aligned} & \dfrac{K{{E}_{2}}}{K{{E}_{1}}}\alpha \dfrac{{{z}^{2}}}{{{n}^{2}}} \\\ & \Rightarrow \dfrac{K{{E}_{2}}}{K{{E}_{1}}}=\dfrac{1}{4} \\\ & \therefore \dfrac{\Delta {{P}_{a}}}{\Delta {{P}_{e}}}=\dfrac{1}{4} \\\ \end{aligned}$
Hence, option b, c are correct options.

Additional information:
Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy. Equivalently, the longer the photon's wavelength, the lower its energy. Photon energy can be expressed using a unit of energy. Among the units commonly used to denote photon energy are the electronvolt (ev) and the joule. Larger units are useful in denoting the energy of photons with higher frequency and higher energy, such as gamma rays.

Note:
We say that an electron or any other particle emits energy when it jumps from higher state orbital to lower state orbital. If an electron or other particle moves from lower orbital to upper orbital, we say that it absorbs energy.