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Ivahew [28]
3 years ago
5

If the threshold wavelength for copper is 2665 A, calculate the maximum kinetic energy of a photoelectron generated by ultraviol

et light of λ 2000 A. Compute the stopping potential.
Physics
1 answer:
Over [174]3 years ago
7 0

Answer:

K.E.=2.48\times 10^{-19}\ J

Explanation:

Using the expression for the photoelectric effect as:

E=h\nu_0+\frac {1}{2}\times m\times v^2

Also, E=\frac {h\times c}{\lambda}

\nu_0=\frac {c}{\lambda_0}

Applying the equation as:

\frac {h\times c}{\lambda}=\frac {hc}{\lambda_0}+\frac {1}{2}\times m\times v^2

Where,

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

\lambda is the wavelength of the light being bombarded

\lambda_0 is the threshold wavelength

\frac {1}{2}\times m\times v^2 is the kinetic energy of the electron emitted.

Given, \lambda=2000\ \dot{A}=2000\times 10^{-10}\ m

\lambda_0=2665\ \dot{A}=2665\times 10^{-10}\ m

Thus, applying values as:

\frac {6.626\times 10^{-34}\times 3\times 10^8}{2000\times 10^{-10}}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2665\times 10^{-10}}+K.E.

K.E.=\frac {6.626\times 10^{-34}\times 3\times 10^8}{2000\times 10^{-10}}-\frac {6.626\times 10^{-34}\times 3\times 10^8}{2665\times 10^{-10}}

K.E.==\frac{19.878}{10^{16}\times \:2000}-\frac{19.878}{10^{16}\times \:2665}

K.E.=2.48\times 10^{-19}\ J

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Explanation:

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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
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Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

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Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

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Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

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Now,

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