Monochromatic light frequency f1 incident on a photocell and the stopping potential is found to be V1. What is the new stopping potential of the-cell, if it is radiated by monochromatic light of frequency f2 ?

Question

Monochromatic light frequency f1 incident on a photocell and the stopping potential is found to be V1. What is the new stopping potential of the-cell, if it is radiated by monochromatic light of frequency f2 ? (A) V1 - ( h / e) ( f2 - f1 )  (B)  V1 + ( h / e) ( f1 + f2 )  (C) V1 - ( h / e) ( f1 + f2 )  (D) V1 + ( h / e) ( f2 - f1 )

Tardigrade Admin · Accepted Answer

Einstein's photoelectric equation can be written as

h f = k + ϕ

but k =

e V_{s}, V_{s}

, being stopping potential

∴ h f = e V_{s} + ϕ

... (i)
So,

h f_{1} = e V_{1} + ϕ

... (ii)
and

h f_{2} = e V_{2} + ϕ

... (iii)
Subtracting Eq. (ii) from Eq. (iii), we get

h (f_{1} - f_{2}) = e (V_{1} - V_{2})

or

e V_{2} = e V_{1} - h (f_{1} - f_{2})

or

V_{2} = V_{1} + \frac{h}{e} (f_{2} - f_{1})

Q. Monochromatic light frequency $f_{1}$ incident on a photocell and the stopping potential is found to be $V_{1}$ . What is the new stopping potential of the-cell, if it is radiated by monochromatic light of frequency $f_{2}$ ?

$V_{1} - \frac{h}{e} (f_{2} - f_{1})$

$V_{1} + \frac{h}{e} (f_{1} + f_{2})$

$V_{1} - \frac{h}{e} (f_{1} + f_{2})$

$V_{1} + \frac{h}{e} (f_{2} - f_{1})$

Q. Monochromatic light frequency f1​ incident on a photocell and the stopping potential is found to be V1​. What is the new stopping potential of the-cell, if it is radiated by monochromatic light of frequency f2​ ?

V1​−eh​(f2​−f1​)

V1​+eh​(f1​+f2​)

V1​−eh​(f1​+f2​)

V1​+eh​(f2​−f1​)

Q. Monochromatic light frequency $f_{1}$ incident on a photocell and the stopping potential is found to be $V_{1}$ . What is the new stopping potential of the-cell, if it is radiated by monochromatic light of frequency $f_{2}$ ?

$V_{1} - \frac{h}{e} (f_{2} - f_{1})$

$V_{1} + \frac{h}{e} (f_{1} + f_{2})$

$V_{1} - \frac{h}{e} (f_{1} + f_{2})$

$V_{1} + \frac{h}{e} (f_{2} - f_{1})$