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

2 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Zero

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Less than 2 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

More than 2 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 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 $V_{s} > 2 V$

Suggest Corrections

Similar questions

Q. In an experiment of photoelectric emission, for incident light of

4000 ˚ A,

the stopping potential is 2V. If the wavelength of incident light is made

3000 ˚ A,

then stopping potential will be

Q. If in a photoelectric cell, the wavelength of incident light is changed from

4000 ˚ A

3600 ˚ A,

then the change in stopping potential will be

Q. A stopping potential of 0.82 volt is required to stop the photo electrons emitted from a metallic surface by light of wavelength

4000 ˚ A .

The stopping potential for wavelength

3000 ˚ A

will be

In a photoelectric experiment for $4000 \circ A$
incident radiation, the potential difference to stop the ejection is 2 V. If the incident light is changed to $3000 \circ A$
, then the potential required to stop the ejection of electrons will be

Q. In a photoelectric effect measurement, the stopping potential for a given metal is found to be

V_{0}

volt when radiation of wavelength

λ_{0}

is used. If radiation of wavelength

2 λ_{0}

is used with the same metal than the stopping potential (in volt) will be:

Join BYJU'S Learning Program

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

The correct option is D More than 2 VPhotoelectric Effect,KEmax=hv−WoThe stopping potential,Vse=KEmaxVse=hv−WoVse=hcλ−WoFrom the above equation, as λ decreases, Vs increases.Hence Vs>2V

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

The correct option is D 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 $V_{s} > 2 V$