Question

In the photoionization of atomic hydrogen when the atom absorbs \[50\,nm\] photon, then the maximum kinetic energy of the emitted electron will be
A. \[7.1\,eV\]
B. \[11.2\,eV\]
C. \[9.2\,eV\]
D. \[19.3\,eV\]

Answer 1

Hint:Find the ionization energy of hydrogen at the ground state. Find the energy of the photon and find the difference to find the kinetic energy of the electron released. The ionization energy of hydrogen at any orbit is inversely proportional to the square of the orbit.

Formula used:
The energy of an electron in \[{n^{th}}\] orbit
\[{E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV\]
where, \[n\] is positive an integer \[1,2,3....\]
\[1eV = {\text{1}}{\text{.60218}} \times {10^{ - 19}}J\]
The energy of a photon of wavelength \[\lambda \] is,
\[E = \dfrac{{hc}}{\lambda }\]
where \[h = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{{\text{1}}{\text{.60218}} \times {{10}^{ - 19}}}}\,eVs\] is the Planck’s constant \[c\] is the velocity of light

Complete step by step answer:
We have given here a photon of wavelength\[50nm\]. We have to find the energy of the released electron. To find that we have to find the ionization energy of hydrogen. Now, the energy of hydrogen at \[{n^{th}}\] orbit is,
\[{E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV\]
So, to escape the electron needs to jump from the first orbit to infinite. Hence ionization energy of hydrogen is,
\[{E_{ion}} = (\dfrac{{ - 13.6}}{{{\infty ^2}}} - \dfrac{{ - 13.6}}{{{1^2}}}) = 13.6eV\]
Now, the energy of the photon is,
\[E = \dfrac{{hc}}{\lambda } = \dfrac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{50 \times {{10}^{ - 9}} \times {\text{1}}{\text{.60218}} \times {{10}^{ - 19}}}}\]
\[\Rightarrow E = 0.248 \times {10^2}eV = 24.8\,eV\]
Now, the energy needed to release the electron is \[13.6\,eV\]. So, rest of the energy will be converted into the kinetic energy of the electron. Hence the kinetic energy of the electron will be,
\[\therefore {E_k} = (24.8 - 13.6) = 11.2\,eV\]
Hence, the kinetic energy of the electron will be \[11.2\,eV\].

Hence, option B is the correct answer.

Note:The ionization energy of an isolated atom is the energy needed to isolate an electron from its outermost shell. Here, the hydrogen atom has only one orbit so we have taken \[n = 1\]. This formula of energy is valid only for a single electron system. For multi electron system the energy of the electron at \[{n^{th}}\] orbit is \[{E_n} = \dfrac{{ - 13.6{Z^2}}}{{{n^2}}}eV\] where, \[Z\] is the atomic number of the element.