In a photoelectric experiment,increasing the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $\rightarrow$ Increasing the intensity of incident light corresponds to an increase in the number of photons incident per unit area per unit time.

Vedclass · Accepted Answer

increases the number of photons incident and the $K.E.$ of the ejected electrons remains unchanged

Vedclass · Answer

(D) $ightarrow$ Increasing the intensity of incident light corresponds to an increase in the number of photons incident per unit area per unit time.$ightarrow$ The kinetic energy $(K.E.)$ of the ejected photoelectrons is determined by the Einstein photoelectric equation: $K.E._{max} = h
u - \phi$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\phi$ is the work function of the metal.$ightarrow$ Since the kinetic energy depends only on the frequency of the incident light and not on its intensity,increasing the intensity does not change the $K.E.$ of the ejected electrons.

In a photoelectric experiment,increasing the intensity of incident light :

Similar Questions

For a given source,with an increase in intensity of incident radiation,the . . . . . .

Photoelectric effect is produced by a point source on a small metal plate. Which of the following curves represents the saturation photoelectric current as a function of the distance between the source and the metal?

$A$ point source causes the photoelectric effect from a small metal plate. Which of the following curves may represent the saturation photocurrent $(I)$ as a function of the distance $(r)$ between the source and the metal?

The figure shows a plot of photo current versus anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement?

Vedclass Test Series

Exam Paper Generator

Online Exam Module