Respuesta :
Explanation:
From first source, kinetic energy ([tex]K.E_{1}[/tex]) ejected is 1 eV and wavelength of light is [tex]\lambda[/tex].
From second source, kinetic energy ([tex]K.E_{2}[/tex]) ejected is 4 eV and wavelength of light is [tex]\frac{\lambda}{2}[/tex].
Relation between work function, wavelength, and kinetic energy is as follows.
K.E = [tex]\frac{hc}{\lambda} - \phi[/tex]
where, h = Plank's constant = [tex]6.63 \times 10^{-34}[/tex] J.s
c = speed of light = [tex]3 \times 10^{8}[/tex] m/s
Also, it is known that 1 eV = [tex]1.6 \times 10^{-19}[/tex] J
Therefore, substituting the values in the above formula as follows.
- From first source,
[tex]K.E_{1}[/tex] = [tex]\frac{hc}{\lambda} - \phi[/tex]
1 eV = [tex]1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi[/tex]
[tex]1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi[/tex] ........... (1)
- From second source,
[tex]K.E_{2}[/tex] = [tex]\frac{hc}{\lambda} - \phi[/tex]
[tex]4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi[/tex]
[tex]6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi[/tex] ........... (2)
Now, divide equation (2) by 2. Therefore, it will become
[tex]{6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}[/tex]
[tex]3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}[/tex] ......... (3)
Now, subtract equation (3) from equation (1), we get the following.
[tex]1.6 \times 10^{-19} = \frac{\phi}{2}[/tex]
[tex]\phi[/tex] = [tex]3.2 \times 10^{-19}[/tex]
= 2 eV
Thus, we can conclude that work function of the metal is 2 eV.
