Question

A photon of wavelength \[4 \times {10^{ - 7}}m\] strikes on the metal surface, the work function of the metal being \[2.13eV\] . Calculate
A) The energy of the photon ( \[eV\] )
B) The kinetic energy of the emission
C) the velocity of the photoelectron. \[(1eV = 1.6020 \times {10^{ - 19}})\]

Answer 1

Hint: The light exhibits both wave and particle nature. The energy of the photon is inversely proportional to its wavelength which means that the shorter the wavelength the more energetic will be the photon and vice versa.

Formula used:
 A) To calculate the energy of the photon, the formula to be used will be;
\[E = hv = \dfrac{{hc}}{\lambda }\] where \[E\] is the energy of the photon, \[\lambda \] is the wavelength of the photon and \[c\]stands for the speed of light in a vacuum, which is \[299792458\] metres per second.
\[h\] denotes Planck’s constant which is \[6.626 \times {10^{ - 34}}({m^2}kg{s^{ - 1}})\] and \[v\] denotes frequency.
B) To calculate the kinetic energy of emission, the formula will be:
\[E = hv - h{v_0}\]
C) The velocity of photoelectron will be calculated by:
\[\dfrac{1}{2}m{v^2}\]

Complete Step by Step Solution:
A) Energy of photon:
After putting the values in \[E = hv = \dfrac{{hc}}{\lambda }\] we will get,
\[ = \dfrac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^{ - 8}}}}{{4 \times {{10}^{ - 7}}}}\]
\[ = 4.97 \times {10^{ - 19}}J\]

B) The kinetic energy of the emission:
After putting the values in \[E = hv - h{v_0}\] we will get,
\[3.1 - 2.13 = \,0.97eV\]

C) The velocity of the photoelectron
After putting the values in \[\dfrac{1}{2}m{v^2}\] we will get,
\[ = 0.97eV = 0.97 \times 1.6 \times {10^{ - 19}}\]
\[\dfrac{1}{2} \times (9.11 \times {10^{ - 31}}){v^2}\]
\[ = \,0.97\, \times \,1.6\, \times \,{10^{ - 19}}\]
\[{v^2} = \,34.1\, \times {10^{10}}\]
\[v = \,5.84\, \times {10^5}m{s^{ - 1}}\]

Note: Photons behave like both a particle and a wave and have a constant velocity that is equal to the speed of light in an empty space. They do not have any mass or rest energy but they do carry energy and momentum which is related to its frequency and wavelength. These photons can be destroyed and can be created when the radiation is absorbed or emitted. They also sometimes exhibit particle-like interactions with the electrons and other atoms which is sometimes seen in the Compton effect.