Answer:
a.1.71 eV
b.0.939 nm
Explanation:
We are given that
Wavelength of light =310 nm=[tex]310\times 10^{-9}[/tex] m
Work function=2.3 eV
Mass of electron=[tex]9.1\times 10^{-31}[/tex]Kg
[tex]\nu=\frac{c}{\lambda}[/tex]
a.We have to find the maximum kinetic energy of ejected electron
[tex]K.E=h\nu-w_0[/tex]
[tex]h={6.626\times 10^{-34}[/tex]
[tex]K.E=\frac{6.626\times 10^{-34}\times 3\times10^8}{310\times10^{-9}\times1.6\times 10^{-19}}-2.3[/tex]
[tex]K.E=\frac{1987.8}{496}-2.3[/tex]
K.E=1.71 eV
Hence, the maximum kinetic energy of ejected electron=1.71 eV
b.Kinetic energy =[tex]\frac{p^2}{2m}[/tex]
p=[tex]\sqrt{2\times 9.1\times 10^{-31}\times 1.71\times1.6\times10^{-19}}[/tex]
p=[tex]7.06\times 10^{-25}[/tex]m-s
We know that de brogile wavelength
[tex]\lambda =\frac{h}{p}[/tex]
[tex]\lambda =\frac{6.626\times10^{-34}}{7.06\times 10^{-25}}[/tex]
[tex]\lambda=0.939\times 10^{-9}[/tex]
[tex]\lambda=0.939 nm[/tex]
Hence, the de-brogile wavelength of ejected electron=0.939 nm.
Answer:
1. 71 eV
0.939 nm
Explanation:
Data:
A. The wavelength of light = [tex]3 * 10^{-9} m[/tex]
work function = 2.3 eV
mass of an electron = [tex]9.1 * 10-^{31} kg[/tex]
the maximum kinetic energy is given by the following equation:
[tex]E_{k} = hv - m_{o} \\ h = 6.626 * 10^{-34}[/tex]
therefore, Ek = [tex]\frac{1987.8}{496} - 2.3\\ = 1.71 eV[/tex]
B. the wavelength is given by:
p = [tex]\sqrt{2*9.1*10^{-31}* 1.71*1.6*10^{-19} }\\= 7.06+ * 10^{-25} ms[/tex]
using the de brogile wavelength equation:
[tex]\lambda = \frac{h}{p}\\ = \frac{6.626*10^{-34} }{7.06*10^{-25} } \\ = 0.939 m[/tex]
