In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

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 ............. $V$.

The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is $\lambda_0$. Two light beams of wavelengths $\frac{\lambda_0}{3}$ and $\frac{\lambda_0}{9}$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photoelectrons (of mass $m$) coming out are respectively $v_1$ and $v_2$,then:

The photoelectric threshold wavelength for a certain metal surface is $3300 \,\mathring{A}$. What is the maximum kinetic energy of the photoelectrons emitted if radiations of wavelength $1100 \,\mathring{A}$ are used?

The threshold frequency for a certain metal is $v_0$. When radiation of frequency $2v_0$ is incident on this metal surface,the maximum velocity of the emitted photoelectrons is $2 \times 10^6 \ m/s$. If radiation of frequency $3v_0$ is incident on the same metal surface,the maximum velocity of the emitted photoelectrons is . . . . . .