Photoelectric Effect

According to Einstein's photoelectric equation, the graph of $\mathrm{K}.\mathrm{E}.$ of the photoelectron emitted from the metal versus the frequency of the incident radiation gives a straight line graph, whose slope

Assertion $\left(\mathrm{A}\right)$- When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{\max }$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{\max }$ increase.

Reason $\left(\mathrm{R}\right)$- Photoelectrons are emitted with speeds ranging from zero to a maximum value, because of the range of frequencies present in the incident light.

The photoelectric threshold wavelength for silver is ${\lambda }_0$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda \left(\lambda <{\lambda }_0\right)$ will be

The work function of a metal is $2.1 \mathrm{eV}$. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surface?

When a certain metal surface is illuminated with light of frequency $\nu$, the stopping potential for photoelectric current is $V_0$. When the same surface is illuminated by the light of frequency $\frac{v}{2}$, the stopping potential is $\frac{V_0}{4}$. The threshold frequency for photoelectric emission is

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

The work function for aluminium is $4.125 \mathrm{eV}$. The cut-off wavelength for the photoelectric effect for aluminium will be

The anode voltage of a photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current $I$ of photocell varies as follows:

The work functions of silver and sodium are $4.6$ and $2.3 \mathrm{eV}$, respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is

Which of the following device is the application of the photoelectric effect?

Find the incorrect statement(s) about the photoelectric effect.

The maximum kinetic energy of the photoelectrons depends only on

The radiation corresponding to $3\to 2$ transition of the hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of $3\times {10}^{-4} \mathrm{T}$. If the radius of the largest circular path followed by these electrons is $10.0 \mathrm{mm}$, the work function of the metal is close to

A metal surface is illuminated by the light of two different wavelengths $248 \mathrm{nm}$ and $310 \mathrm{nm}$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$, respectively. If the ratio $u_1:u_2=2:1$ and $hc=1240 \mathrm{eV} \mathrm{nm}$, the work function of the metal is nearly

Assertion- If light continuously falls on a metal surface, then the emission of electrons will stop after some time.

Reason- We cannot extract all the electrons of a metal.

Assertion- If the wavelength of incident photons is halved, then the maximum kinetic energy of photoelectrons will become two times.

Reason- Energy of the photon$\propto \frac{1}{\lambda }$.

Assertion- The red light is just able to emit photoelectrons from a metal surface, then violet light cannot emit the photoelectrons from that metal.

Reason- The energy of violet light photons is more than red light photons.

Assertion: If distance of the point source is increased from the photoelectric plate, then stopping potential will remain unchanged.

Reason: Saturation current will decrease.

A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_0$. If an electron having maximum kinetic energy enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of the radius $r$, then the radius $r$ is given by, (in the usual notation)

When photons of energy $4.25 \mathrm{eV}$ strike the surface of a metal $\mathrm{A}$, the ejected photoelectron have maximum kinetic energy $T_{\mathrm{A}}$ expressed in $\mathrm{eV}$ and de-Broglie wavelength ${\lambda }_{\mathrm{A}}$. The maximum kinetic energy of photoelectron liberated from another metal $\mathrm{B}$ by photons of energy $4.70 \mathrm{eV}$ is $T_{\mathrm{B}}=\left(T_{\mathrm{A}}-1.50\right) \mathrm{eV}$. If the de-Broglie wavelength of these photoelectrons is ${\lambda }_{\mathrm{B}}=2{\lambda }_{\mathrm{A}}$, then