Respuesta :
Answer:
Velocity of electron will be [tex]v=0.986\times 10^6m/sec[/tex]
Explanation:
We have given work function of metal [tex]\Phi =1.05eV=1.05\times 1.6\times 10^{-19}J=1.68\times 10^{-19}J[/tex]
Wavelength of the light [tex]\lambda =324nm=324\times 10^{-9}m[/tex]
So energy is given by [tex]E=\frac{hc}{\lambda }[/tex], here h is plank's constant and c is speed of light
So [tex]E=\frac{6.6\times 10^{-34}\times3\times 10^8}{324\times 10^{-9} }=6.11\times 10^{-19}j[/tex]
For a metal we know that [tex]E=\Phi +KE_{MAX}[/tex]
So [tex]KE_{MAX}=E-\Phi =6.11\times 10^{-19}-1.68\times 10^{-19}=4.43\times 10^{-19}[/tex]
Now kinetic energy is given by [tex]KE=\frac{1}{2}mv^2[/tex]
[tex]4.43\times 10^{-19}=\frac{1}{2}\times 9.11\times 10^{-31}v^2[/tex]
[tex]v=0.986\times 10^6m/sec[/tex]
So velocity of electron will be [tex]v=0.986\times 10^6m/sec[/tex]
The maximum velocity is obtained by charged particles when electrons in a material owing to an electric field. The maximum velocity of the photoelectron will be 0.986 × 10⁶m/sec.
What is the maximum velocity of the photoelectron?
The average velocity is obtained by charged particles when electrons in a material owing to an electric field.
The work function of the metal is given 1.05 eV.
WF = 1.05 e
WF = 1.05 ×1.6 ×10⁻¹⁹eV
WF = 1.68 ×10⁻¹⁹ J
λ= 324 nm= 324 ×10⁻⁹ m
[tex]E = \frac{hc}{\lambda}[/tex]
[tex]E = \frac{6.6\times10^{-34}}{324\times10^{-9}}[/tex]
[tex]E = 6.11 \times 10^{-19}[/tex]
For a metal
E = ∅+KE
KE = E - ∅
KE = 6.11×10⁻¹⁹ - 1.68×10⁻¹⁹
KE = 4.43 × 10⁻¹⁹ J
[tex]\rm{KE = \frac{1}{2} mV^{2} }[/tex]
[tex]\rm{V^{2} }=\frac{2KE}{m}[/tex]
[tex]\rm{V= \sqrt{\frac{2KE}{m}}[/tex]
[tex]\rm{V= \sqrt{\frac{2\times 4.43 \times10^{-19}}{9.11\times 10^{-31}}}[/tex]
[tex]\rm{V=0.986\times 10^6}[/tex] m/sec.
Hence the maximum velocity of the photoelectron will be 0.986 × 10⁶m/sec.
To learn more about the maximum velocity of the photoelectrons refer
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