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Question: Threshold wavelength for photoelectric emission from a metal surface is 5200 \({A^0}\) Photoelectron...

Threshold wavelength for photoelectric emission from a metal surface is 5200 A0{A^0} Photoelectrons will be emitted when this surface is illuminated with monochromatic radiation from:
A. 1 W IR lamp
B. 50 W UV lamp
C. 50 W IR lamp
D. 10 W IR lamp

Explanation

Solution

Hint : When light(photons) falls on a surface of metal then electrons gain some energy from it and get emitted with some kinetic energy .This phenomena is called photo electric effect. Also when the light falls on the surface of the metal like sodium, rubidium etc. then their electrons start emitting and this phenomenon is known as photoelectric emission. We will use definition to solve the question accordingly.

Complete step-by-step answer:
The condition for photoelectric effect is that the light must have energy greater than a minimum energy (called threshold energy) then only it will be able to emit electrons.
Thus, its equation is given by
K.E=E−E0 K.E=hv−hvo K.E=hcλ−hcλ0  K.E = E - {E_0} \\\ K.E = hv - h{v_o} \\\ K.E = \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} \\\
Where
K.E. is the kinetic energy, h is the planck's constant, c is the velocity of light and λ\lambda is the wavelength.
Using the above equation to solve the question
Since, K.E>0K.E > 0 for photoelectric effect.
Therefore
⇒hcλ−hcλ0>0 ⇒hcλ>hcλ0 ⇒λ<λ0 ⇒λ<5200A0 or 520nm  \Rightarrow \dfrac{{hc}}{\lambda } - \dfrac{{hc}}{{{\lambda _0}}} > 0 \\\ \Rightarrow \dfrac{{hc}}{\lambda } > \dfrac{{hc}}{{{\lambda _0}}} \\\ \Rightarrow \lambda < {\lambda _0} \\\ \Rightarrow \lambda < 5200{A^0}{\text{ or }}520nm \\\
Since our visible range is from 400nm to 700nm, the UV rays range is less than 400nm and the infrared range is more than 700nm .Therefore, UV rays can be used to illuminate the surface of this metal.
Hence the correct option is B.

Note : We can see that here in options the power of light is also given which is directly linked with the intensity of light. It does not have to do anything with the energy of emitted electrons in photoelectric effect. It is only related to the number of electrons emitted if power is more, intensity will be more thus more electrons will be emitted .