In a photoelectric experiment,if the wavelength of incident radiation is reduced from $6000 \ \mathring{A}$ to $4000 \ \mathring{A}$ while keeping the intensity of radiation constant,then:

An electron and a proton are separated by a large distance. The electron starts approaching the proton with an initial kinetic energy of $3 \, eV$. The proton captures the electron and forms a hydrogen atom in the second excited state. The resulting photon is incident on a photosensitive metal with a threshold wavelength of $4000 \, Å$. What is the maximum kinetic energy of the emitted photoelectron? (In $eV$)

When light of a given wavelength is incident on a metallic surface,the minimum potential needed to stop the emitted photoelectrons is $6.0 \ V$. This potential drops to $0.6 \ V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal,respectively? $\left[\text{Take } hc = 1.24 \times 10^{-6} \ J \ m\right]$

In an experiment on the photoelectric effect,the wavelength of the incident radiation is $\lambda$. The wavelength of the incident radiation is reduced to $\frac{1}{3}$ of the initial value,and the maximum kinetic energy of the photoelectron is observed to be $n$ times the previous value. The threshold wavelength for the metal plate is:

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectron emitted from the surface has mass $m$ and de-Broglie wavelength $\lambda_{d}$,then

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is: