The frequency of the incident light falling o | vedclass

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

Question

(C) According to Einstein's photoelectric equation: $K_{\max} = h
u - W_0$,where $K_{\max}$ is the maximum kinetic energy,$h$ is Planck's constant,$\

Vedclass · Accepted Answer

More than doubled

Vedclass · Answer

(C) According to Einstein's photoelectric equation: $K_{\max} = h
u - W_0$,where $K_{\max}$ is the maximum kinetic energy,$h$ is Planck's constant,$
u$ is the frequency of incident light,and $W_0$ is the work function of the metal.Let the initial frequency be $
u$. The initial kinetic energy is $K_1 = h
u - W_0$.When the frequency is doubled,the new frequency becomes $2
u$. The new kinetic energy is $K_2 = h(2
u) - W_0 = 2h
u - W_0$.Comparing $K_2$ with $2K_1$:$2K_1 = 2(h
u - W_0) = 2h
u - 2W_0$.Since $W_0 > 0$,it follows that $2h
u - W_0 > 2h
u - 2W_0$.Therefore,$K_2 > 2K_1$. The kinetic energy becomes more than doubled.

The frequency of the incident light falling on a photosensitive metal plate is doubled. The kinetic energy of the emitted photoelectrons is:

Similar Questions

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

The following graphs show the variation of stopping potential $(V_s)$ corresponding to the frequency of incident radiation $(f)$ for a given metal. The correct variation is shown in graph ($f_0 =$ Threshold frequency):

Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

The work function of a substance is $4.0 \,eV$. The longest wavelength of light that can cause photoelectron emission from this substance is approximately ......... $nm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module