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Question: Threshold frequency of a metal is \( 5 \times \,{10^{13}}{\sec ^{ - 1}} \) upon which \( 1 \times \,...

Threshold frequency of a metal is 5× 1013sec⁡−15 \times \,{10^{13}}{\sec ^{ - 1}} upon which 1× 1014sec⁡−11 \times \,{10^{14}}{\sec ^{ - 1}} frequency light focused, then maximum kinetic energy of emitted electrons?
(A) 3.3× 10−213.3 \times \,{10^{ - 21}}
(B) 3.3× 10−203.3 \times \,{10^{ - 20}}
(C) 6.6× 10−216.6 \times \,{10^{ - 21}}
(D) 6.6× 10−206.6 \times \,{10^{ - 20}}

Explanation

Solution

Hint : Threshold frequency of a metal refers to the frequency of light that causes an ejection of electrons from a metal surface. To solve this question, we must recall the principle of conservation of energy by which we can obtain the kinetic energy as the difference of the incident energy and the threshold energy.

Complete Step By Step Answer:
We know that the threshold frequency of a metal refers to the minimum frequency of light that will cause an electron to eject from the surface of the metal. The frequency of light below threshold frequency will not eject an electron. The value of frequency at the threshold frequency will eject an electron with no kinetic energy.
According to the principle of conservation of energy, it states that energy can neither be created, nor be destroyed but it transforms from one form to another. Einstein's photoelectric equation is based on the law of conservation of energy. The energy carried by each particle of the light i.e. photon depends on the frequency of the incident light.
We know that,
Kinetic energy = 12mv2\dfrac{1}{2}m{v^2}
Where,
mm = mass of the object
vv =velocity of the object
Also, kinetic energy (K.E)(K.E) = h(ν−νo)h(\nu - {\nu _o})
Where,
hh =Planck’s constant
ν\nu = frequency of the incident light
νo{\nu _o} = threshold frequency
In the given question,
Frequency of the incident light (ν)(\nu ) = 1× 1014sec⁡−11 \times \,{10^{14}}{\sec ^{ - 1}}
Threshold frequency (νo)({\nu _o}) = 5× 1013sec⁡−15 \times \,{10^{13}}{\sec ^{ - 1}}
Value of Planck’s constant (h)(h) = 6.626 ×10−34Js6.626\, \times {10^{ - 34}}Js
We know,
Kinetic energy (K.E)(K.E) = h(ν−νo)h(\nu - {\nu _o})
Substituting all the values, we get
K.E=(6.626 × 10−34Js) (1×1014s−1 −5 × 1013s−1)K.E = (6.626\, \times \,{10^{ - 34}}Js)\,(1 \times {10^{14}}{s^{ - 1}}\, - 5\, \times \,{10^{13}}{s^{ - 1}})
K.E=(6.626 × 10−34Js) ( 5 × 1013s−1)K.E = (6.626\, \times \,{10^{ - 34}}Js)\,(\,5\, \times \,{10^{13}}{s^{ - 1}})
K.E=3.313× 10−20JK.E = 3.313 \times \,{10^{ - 20}}J
The maximum kinetic energy of the emitted electrons is 3.313× 10−20J3.313 \times \,{10^{ - 20}}J
Hence, the correct answer is option (B).

Note :
Always remember when the ejection of electrons occurs at the threshold frequency then, the value of kinetic energy will be zero. Incident frequency below the threshold frequency will not eject an electron. Also, when the incident light is above the threshold frequency will eject an electron with some kinetic energy. This phenomenon is known as the photoelectric effect.