Respuesta :
Explanation:
According to the Einstein law, it is known that
[tex]h \times \nu = \phi + \frac{1}{2} mv^{2}[/tex]
where, h = energy of light
[tex]\phi[/tex] = work function
[tex]m v^{2}[/tex] = kinetic energy of electron
It is given that the value of [tex]h \nu[/tex] is [tex]7.2 \times 10^{-19} J[/tex]. And,
1 eV = [tex]1.6 \times 10^{-19} J[/tex]
Here, [tex]\phi[/tex] for titanium is 4.33 eV
= [tex](4.33 \times 1.6 \times 10^{-19})[/tex] J
= [tex]6.928 \times 10^{-19}[/tex] J
(a) First of all, kinetic energy will be calculated as follows.
[tex]\frac{1}{2}mv^{2} = h \nu - \phi[/tex]
= [tex](7.2 \times 10^{-19} - 6.92 \times 10^{-19})[/tex] J
= [tex]0.272 \times 10^{-19}[/tex] J
It is known that mass of electrons is equal to [tex]9.109 \times 10^{-31} kg[/tex].
Therefore, [tex]mv^{2} = 0.544 \times 10^{-19} J[/tex]
and, [tex](mv)^{2} = 9.109 \times 0.544 \times 10^{-19} \times 10^{-31}[/tex]
= [tex]4.955 \times 10^{-50}[/tex]
mv = [tex]2.225 \times 10^{-25}[/tex]
Now, the relation between wavelength and mv is as follows.
[tex]\lambda = \frac{6.626 \times 10^{-34}}{2.225 \times 10^{-25}}[/tex]
= [tex]2.98 \times 10^{-9} m[/tex]
Therefore, the wavelength of the ejected electrons is [tex]2.98 \times 10^{-9} m[/tex].
(b) It is known that relation between energy and wavelength is as follows.
E = [tex]h \nu = \frac{hc}{\lambda}[/tex]
[tex]\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{7.2 \times 10^{-19}}[/tex]
= [tex]\frac{6.626 \times 3 \times 10^{-26}}{7.2 \times 10^{-19}}[/tex]
= [tex]2.76 \times 10^{-7} m[/tex]
Hence, the wavelength of the ejected electrons is [tex]2.76 \times 10^{-7} m[/tex].
(c) For iron surface, [tex]\phi = 4.7 eV[/tex]
= [tex](4.7 \times 1.6 \times 10^{-19})[/tex] J
= [tex]7.52 \times 10^{-19}[/tex] J
Here, the value of [tex]\phi[/tex] is more than the value of UV light source. Hence, we need a shorter wavelength light as we know that,
[tex]E \propto \frac{1}{\lambda}[/tex]
Therefore, lesser will be the wavelength higher will be the energy.
