The work function of a certain metal is $3.31 \times 10^{-19} \,J$. The maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 \text{ Å}$ is (Given: $h = 6.62 \times 10^{-34} \,J \cdot s$,$c = 3 \times 10^8 \,m/s$,$e = 1.6 \times 10^{-19} \,C$) (in $\text{ eV}$)

Light of two different frequencies whose photons have energies $1 \text{ eV}$ and $2.5 \text{ eV}$ respectively,successively illuminates a metal of work function $0.5 \text{ eV}$. The ratio of the maximum kinetic energies of the emitted electrons will be:

For the photoelectric effect with incident photon wavelength $\lambda$,the stopping potential is $V_0$. Identify the correct variation$(s)$ of $V_0$ with $\lambda$ and $1/\lambda$.

If the electron in a hydrogen atom jumps from the third Bohr orbit to the ground state directly and the difference between the energies of the two states is radiated in the form of photons. If the work function of the material is $4.1 \text{ eV}$, then the stopping potential is nearly:
$\left[\text{Energy of electron in } n^{\text{th}} \text{ orbit} = \frac{-13.6}{n^2} \text{ eV}\right]$ (in $\text{ V}$)

When light of frequency $v$ is incident on a metal,photoelectrons are emitted. This is because...

In a photoelectric cell,the wavelength of incident light is changed from $4000 \, \mathring{A}$ to $3600 \, \mathring{A}$. The change in stopping potential will be ............. $V$.