Q.

Consider the experimental set-up of verification of photoelectric effect shown in the figure. The voltage across the electrodes is measured with the help of an ideal voltmeter and which can be varied by moving jockey J on the potentiometer wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is  2Ω. The resistance of 100 cm along potentiometer wire is  8Ω.

Question Image

The photocurrent is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50cm2  at separation 0.5 mm are used in the vacuum tube. Photocurrent in the circuit is very small, so we can treat potentiometer circuit an independent circuit. Using the information in the table here, choose correct alternative(s).

Light

1

Violet

2

Blue

3

Green

4

Yellow

5

Orange

6

Red

       λ(in A)4000-45004500-50005000-55005500-60006000-65006500-7000

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a

The number of electrons appearing on the surface of the cathode plate, when the jockey is connected at the end P of the potentiometer wire is  n=8.85×109, even if no radiation falls on the cathode

b

It is found that on incidence of a light (of power  4×106W), current ( 20μA) remains almost unchanged even when the jockey is moved from the end P to the middle point of the potentiometer wire. Assuming all the incident photons eject electron, and the light is orange.

c

Red may or may not reject photoelectrons but violet must eject them

d

When orange light falls on the anode plate, the ammeter reading remains zero till jockey is moved from the end P to the middle point of the wire PQ. There after the deflection is recorded in the ammeter. Average of minimum and maximum kinetic energies of the emitted electron is 6 eV.

answer is A, B, C.

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Detailed Solution

Regarding option (a)
Treating the arrangement of plates as parallel plate capacitor, charge on capacitor will be  Q=CV
       ne=ε0AdV      n=8.85×1012×100.5×103×1.6×1019×16     n=8.85×109  
Regarding option (b)
 Incident power can be written as
 Where, n = number of photons incident per unit time.
 Also, saturation current,
 I = ne  P=lhceλλ=lhceP
=(2×106)(6.6×1034)(3×108)(4×106)(1.6×1019)=9.91.6×107m=6187A
 Which corresponds to given range of orange
Regarding option (c)
 As seen in option (b), orange is ejecting but as wavelength range it may or greater than that of orange it may or may not eject but as wavelength of violet is lesser is must.
Regarding option (d)
 Stopping potential,  Vs= 8 V
KEmax=8 eV while KEmin=0,so  
 Average energy will be 4 eV.

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