In a photoelectric experiment the stopping potential for the incident light of wavelength 4000 mathringA is 2 V. If the wavelength be changed to 3000 mathringA , the stopping potential will be

Question

In a photoelectric experiment the stopping potential for the incident light of wavelength 4000 mathringA is 2 V. If the wavelength be changed to 3000 mathringA , the stopping potential will be (A) 2 V (B) zero (C) less than 2 V (D) more than 2 V

Tardigrade Admin · Accepted Answer

Photoelectric Effect, K E max = h v - W o The stopping potential, V s e = K E max V s e = h v - W o V s e =( h c /λ)- W o From the above equation, as λ decreases, V s increases. Hence V s > 2 V

Q. In a photoelectric experiment the stopping potential for the incident light of wavelength $4000 \overset{˚}{A}$ is 2 V. If the wavelength be changed to $3000 \overset{˚}{A}$ , the stopping potential will be

2 V

zero

less than 2 V

more than 2 V

Photoelectric Effect,
$K E_{m a x} = h v - W_{o}$
The stopping potential,
$V_{s} e = K E_{m a x}$
$V_{s} e = h v - W_{o}$
$V_{s} e = \frac{h c}{λ} - W_{o}$
From the above equation, as $λ$ decreases, $V_{s}$ increases. Hence $V_{s} > 2 V$

Q. In a photoelectric experiment the stopping potential for the incident light of wavelength 4000A˚ is 2 V. If the wavelength be changed to 3000A˚, the stopping potential will be

2 V

zero

less than 2 V

more than 2 V

Photoelectric Effect, KEmax​=hv−Wo​ The stopping potential, Vs​e=KEmax​ Vs​e=hv−Wo​ Vs​e=λhc​−Wo​ From the above equation, as λ decreases, Vs​ increases. Hence Vs​>2V

Q. In a photoelectric experiment the stopping potential for the incident light of wavelength $4000 \overset{˚}{A}$ is 2 V. If the wavelength be changed to $3000 \overset{˚}{A}$ , the stopping potential will be

Photoelectric Effect,
$K E_{m a x} = h v - W_{o}$
The stopping potential,
$V_{s} e = K E_{m a x}$
$V_{s} e = h v - W_{o}$
$V_{s} e = \frac{h c}{λ} - W_{o}$
From the above equation, as $λ$ decreases, $V_{s}$ increases. Hence $V_{s} > 2 V$