If the wavelength of incident light is decreased from $4000 \ \mathring{A}$ to $3600 \ \mathring{A}$,then the change in stopping potential will be ............. $V$.

$A$ beam of electromagnetic radiation of intensity $6.4 \times 10^{-5} \; W/cm^{2}$ is comprised of wavelength $\lambda = 310 \; nm$. It falls normally on a metal surface (work function $\varphi = 2 \; eV$) of surface area $1 \; cm^{2}$. If one in $10^{3}$ photons ejects an electron, the total number of electrons ejected in $1 \; s$ is $10^{x}$. Then $x$ is: $(hc = 1240 \; eV \cdot nm, 1 \; eV = 1.6 \times 10^{-19} \; J)$

The study of photoelectric effect is useful in understanding

$A$ beam of light of wavelength $400\,nm$ and power $1.55\,mW$ is directed at the cathode of a photoelectric cell. If only $10\%$ of the incident photons effectively produce photoelectrons,then find the current due to these electrons in $\mu A$. (Given: $hc = 1240\,eV\cdot nm$,$e = 1.6 \times 10^{-19}\,C$)

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

The stopping potential for photoelectrons: