Answer
77.7k+ views
Hint: Use Einstein’s photoelectric equation to find the maximum kinetic energy. Apply the given conditions in the question to obtain the expression for work function.
Complete step by step solution:
Let us assume that W is the work function material on the surface
According to Einstein’s Photoelectric Equation,
\[K{E_{\max }} = \dfrac{{hc}}{\lambda } - W\]------ (1)
Now, when \[\lambda \] becomes \[\dfrac{\lambda }{2}\] in the second case, the equation changes into,
\[K{E_{2\max }} = \dfrac{{hc}}{\lambda } - W = \dfrac{{2hc}}{\lambda } - W\] -------- (2)
It is given in the question that the maximum kinetic energy of photons emitted in the second case is 3 times the first case.
Therefore,
\[K{E_{2\max }} = 3 \times K{E_{\max }}\]
Substituting $K{E_{\max }}$ value from equation (1), we get
\[K{E_{2\max }} = 3 \times (\dfrac{{hc}}{\lambda } - W)\]
\[ \Rightarrow \dfrac{{2hc}}{\lambda } - W = \dfrac{{3hc}}{\lambda } - 3W\]
\[ \Rightarrow 2W = \dfrac{{hc}}{\lambda }\]
\[ \therefore W = \dfrac{{hc}}{{2\lambda }}\]
Hence, the work function W is given as \[\dfrac{{hc}}{{2\lambda }}\].
Option (B) is the right answer.
Note: The photoelectric effect is the emission of electrons, when electromagnetic radiation hits the material. Electrons emitted are called photoelectrons. Above the specified frequency value, the strength of the photoelectric current depends on the light radiation intensity. Any frequency value below the specified value will be unable to produce photoelectrons or photoelectric current.
Albert Einstein said that a beam of light will not propagate as waves rather it propagates as discrete packets of waves called as photons. It stated that light is a collection of quantum radiation. Einstein combined his theory with the photoelectric concept put forward by Max Planck. He states that each
Complete step by step solution:
Let us assume that W is the work function material on the surface
According to Einstein’s Photoelectric Equation,
\[K{E_{\max }} = \dfrac{{hc}}{\lambda } - W\]------ (1)
Now, when \[\lambda \] becomes \[\dfrac{\lambda }{2}\] in the second case, the equation changes into,
\[K{E_{2\max }} = \dfrac{{hc}}{\lambda } - W = \dfrac{{2hc}}{\lambda } - W\] -------- (2)
It is given in the question that the maximum kinetic energy of photons emitted in the second case is 3 times the first case.
Therefore,
\[K{E_{2\max }} = 3 \times K{E_{\max }}\]
Substituting $K{E_{\max }}$ value from equation (1), we get
\[K{E_{2\max }} = 3 \times (\dfrac{{hc}}{\lambda } - W)\]
\[ \Rightarrow \dfrac{{2hc}}{\lambda } - W = \dfrac{{3hc}}{\lambda } - 3W\]
\[ \Rightarrow 2W = \dfrac{{hc}}{\lambda }\]
\[ \therefore W = \dfrac{{hc}}{{2\lambda }}\]
Hence, the work function W is given as \[\dfrac{{hc}}{{2\lambda }}\].
Option (B) is the right answer.
Note: The photoelectric effect is the emission of electrons, when electromagnetic radiation hits the material. Electrons emitted are called photoelectrons. Above the specified frequency value, the strength of the photoelectric current depends on the light radiation intensity. Any frequency value below the specified value will be unable to produce photoelectrons or photoelectric current.
Albert Einstein said that a beam of light will not propagate as waves rather it propagates as discrete packets of waves called as photons. It stated that light is a collection of quantum radiation. Einstein combined his theory with the photoelectric concept put forward by Max Planck. He states that each
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