When a photon with energy $8\ eV$ is incident on a metal surface having a threshold frequency of $1.6 \times 10^{15}\ Hz$,the maximum kinetic energy of the emitted photoelectrons is ............ $eV$. (Given: $h = 6.6 \times 10^{-34}\ J\cdot s$,$1\ eV = 1.6 \times 10^{-19}\ J$)

When a certain photosensitive surface is illuminated with monochromatic light of frequency $v$,the stopping potential for the photocurrent is $-V_0/2$. When the surface is illuminated by monochromatic light of frequency $v/2$,the stopping potential is $-V_0$. The threshold frequency for photoelectric emission is

The correct graph between the maximum kinetic energy $(K_{\max})$ of a photoelectron and the inverse of the wavelength $(1/\lambda)$ of the incident radiation is given by the curve:

The work functions of Aluminium and Gold are $4.1\,eV$ and $5.1\,eV$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is $.......$

The photoelectric cut-off voltage in a certain experiment is $1.5 \text{ V}$. The maximum kinetic energy of photoelectrons emitted will be . . . . . . .

Light of wavelength $\lambda$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy $E$. If the kinetic energy is to be increased to $2E$,the wavelength must be changed to $\lambda'$ where