Answer:
[tex]9.56\times 10^5m/s[/tex]
Explanation:
We are given that
Wavelength=[tex]\lambda=245nm=245\times 10^{-9} m[/tex]
[tex]1nm=10^{-9} m[/tex]
Work function=[tex]w_0=5.1eV=5.1\times 1.6\times 10^{-19} C[/tex]
[tex]1e=1.6\times 10^{-19} C[/tex]
We have to find the maximum speed of photo-electrons emitted from this surface.
Mass of electron,m=[tex]9.1\times 10^{-31} kg[/tex]
We know that
[tex]\frac{1}{2}mv^2=\mid\frac{hc}{\lambda}-w_0\mid[/tex]
Where [tex]h=6.63\times 10^{-34}[/tex]
[tex]c=3\times 10^8m/s[/tex]
Using the formula
[tex]\frac{1}{2}\times 9.1\times 10^{-31}v^2=\mid \frac{6.63\times 10^{-34}\times 3\times 10^8}{245\times 10^{-9}}-5.1\times 1.6\times 10^{-19}\mid[/tex]
[tex]\frac{1}{2}\times 9.1\times 10^{-31}v^2=4.16\times 10^{-21}[/tex]
[tex]v^2=\frac{4.16\times 10^{-21}\times 2}{9.1\times 10^{-31}}[/tex]
[tex]v=\sqrt{\frac{4.16\times 10^{-21}\times 2}{9.1\times 10^{-31}}}=9.56\times 10^5m/s[/tex]
The maximum speed of the electron emitted from the metal surface is [tex]1.1 \times 10^5 \ m/s[/tex].
The given parameters;
The energy of the incident light is calculated as follows;
[tex]E = hf\\\\E = \frac{hc}{\lambda} \\\\E = \frac{(6.626 \times 10^{-34} ) \times 3\times 10^8}{245 \times 10^{-9}} \\\\ E= 8.11 \times 10^{-19} \ J[/tex]
The kinetic energy of the electron emitted from the metal surface is calculated as follows;
[tex]K.E = E - \phi \\\\K.E = (8.11 \times 10^{-19}) \ - \ (5.1 \times 1.6 \times 10^{-19})\\\\K.E = -5 \times 10^{-21} \ J\\\\|K.E| = 5 \times 10^{-21} \ J[/tex]
The speed of the electron is calculated as;
[tex]\frac{1}{2} mv^2 = K.E\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2(5\times 10^{-21})}{9.11 \times 10^{-31}} } \\\\v = 1.1 \times 10^5 \ m/s[/tex]
Thus, the maximum speed of the electron emitted from the metal surface is [tex]1.1 \times 10^5 \ m/s[/tex].
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