In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

$A$ material $A$ exhibits the photoelectric effect. Its work function is $2.5 \ eV$ and the threshold wavelength is $\lambda$. Another material $B$ has a work function of $5 \ eV$. Find the threshold wavelength required to produce the photoelectric effect in $B$.

In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

The surface of a certain metal is first illuminated with light of wavelength $\lambda_1 = 350 \ nm$ and then with light of wavelength $\lambda_2 = 540 \ nm$. It is found that the maximum speed of the photoelectrons in the two cases differs by a factor of $2$. The work function of the metal (in $eV$) is close to: (Energy of photon $= \frac{1240}{\lambda \text{ (in } nm)} \ eV$)

The electron in the hydrogen atom jumps from excited state $(n = 3)$ to its ground state $(n = 1)$ and the photons thus emitted irradiate a photosensitive material. If the work function of the material is $5.1 \, eV$,the stopping potential is estimated to be (the energy of the electron in $n^{th}$ state $E_n = \frac{-13.6}{n^2} \, eV$) (in $, V$)