Q 11 :    

Which figure shows the correct variation of applied potential difference (V) with photoelectric current (i) at two different intensities of light (I1<I2) of same wavelengths?  [2024]

 

 

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  •  

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(1)

Saturation current depends on intensity of incident light.
Also, according to question, both have same wavelength.
Hence stopping potential will remain same, it means KEmax will be same.
Since I2>I1, hence saturation current corresponding to I2 will be greater than that corresponding to I1.

 



Q 12 :    

Given below are two statements:

Statement I: Figure shows the variation of stopping potential with frequency (ν) for the two photosensitive materials M1and M2. The slope gives value of he, where h is Planck’s constant, e is the charge of an electron.

Statement II: M2 will emit photoelectrons of greater kinetic energy for the incident radiation having the same frequency.

In the light of the above statements, choose the most appropriate answer from the options given below.                 [2024]

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are incorrect

     

  • Both Statement I and Statement II are correct

     

  • Statement I is correct and Statement II is incorrect

     

(4)

eVs=hν+ϕ  Vs=heν+ϕe

slope = he

Work function of M2 is higher than M1 so kinetic energy of emitted electron is less for M2



Q 13 :    

For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (v) of the incident photons as shown in the figure. The slope of the graph gives             [2024]

  • Ratio of Planck’s constant to electric charge

     

  • Work function of the metal

     

  • Charge of electron

     

  • Planck’s constant

     

(4)

We know that

E=K.E.+ϕ0

hν=K.E.+ϕ0

K.E=hν-ϕ0

Slope of kinetic energy versus frequency curve will be equal to slope h.