An experimental setup for verification of photoelectric effect is shown in the diagram. The voltage across the electrodes is measured with the help of an ideal voltmeter, and it can be varied by moving jockey J on the potentiometer wire. The battery used in potentiometric circuit is of 20 V and its internal resistance is 2 Ω. The resistance of 100 cm long potentiometer wire is 8 Ω.
The photo current is measured with the help of an ideal ammeter.
The wavelength of various colours is as follows:
Violet:4000 ˚A−5000 ˚A
Blue:4500 ˚A−5000 ˚A
Green:5000 ˚A−5500 ˚A
Yellow:5500 ˚A−6000 ˚A
Orange:6000 ˚A−6500 ˚A
Red:6500 ˚A−7000 ˚A
It is found that the ammeter current remains unchanged (2 μA) even when the jockey is moved from the end P to the middle point of the potentiometer wire. Assuming one photon ejects one electron and the power of the light incident is 4×10−6 W, the colour of the incident light is:
An experimental setup for verification of photoelectric effect is shown in the diagram. The voltage across the electrodes is measured with the help of an ideal voltmeter, and it can be varied by moving jockey J on the potentiometer wire. The battery used in potentiometric circuit is of 20 V and its internal resistance is 2 Ω. The resistance of 100 cm long potentiometer wire is 8 Ω.
The photo current is measured with the help of an ideal ammeter.
The wavelength of various colours is as follows:
Violet:4000 ˚A−5000 ˚A
Blue:4500 ˚A−5000 ˚A
Green:5000 ˚A−5500 ˚A
Yellow:5500 ˚A−6000 ˚A
Orange:6000 ˚A−6500 ˚A
Red:6500 ˚A−7000 ˚A
It is found that the ammeter current remains unchanged (2 μA) even when the jockey is moved from the end P to the middle point of the potentiometer wire. Assuming one photon ejects one electron and the power of the light incident is 4×10−6 W, the colour of the incident light is:
The correct option is D Orange
The energy of one photon is given by,
E=hcλ
If n number of photons strike per sec, the rate of incident energy will be,
P=nhcλ ...(1)
Also, the number of ejected electrons is the same as the number of incident photons per unit time.
Thus, I=ne ...(2)
From (1) and (2),
P=hcIeλ
Given,
h=6.6×10−34 J-s, c=3×108 m/s, I=2×10−6A
⇒P=(2×10−6)×(6.6×10−34)×(3×108)1.6×10−19×λ
⇒4×10−6=(6.6×2×3)×10−32(1.6×10−19×λ)
⇒λ=6.6×6×10−324×10−6×1.6×10−19=6.1875×10−7 m
⇒λ=6187.5×10−10 m=6187.5 ˚A
Thus, the incident light is of Orange colour.
Hence, option (D) is correct.
The energy of one photon is given by,
E=hcλ
If n number of photons strike per sec, the rate of incident energy will be,
P=nhcλ ...(1)
Also, the number of ejected electrons is the same as the number of incident photons per unit time.
Thus, I=ne ...(2)
From (1) and (2),
P=hcIeλ
Given,
h=6.6×10−34 J-s, c=3×108 m/s, I=2×10−6A
⇒P=(2×10−6)×(6.6×10−34)×(3×108)1.6×10−19×λ
⇒4×10−6=(6.6×2×3)×10−32(1.6×10−19×λ)
⇒λ=6.6×6×10−324×10−6×1.6×10−19=6.1875×10−7 m
⇒λ=6187.5×10−10 m=6187.5 ˚A
Thus, the incident light is of Orange colour.
Hence, option (D) is correct.
