If ultraviolet radiation of $6.2 eV$ falls on an aluminium surface,then the kinetic energy of the fastest emitted electron is (work-function $= 4.2 eV$)

$A$ monochromatic light of wavelength $\lambda$ ejects photoelectrons from a metal surface with work function $\phi = 2.4 \text{ eV}$. These photoelectrons are made to collide with hydrogen atoms in the ground state. The maximum value of $\lambda$ for which the hydrogen atom may be ionized is [take $hc = 1240 \text{ eV-nm}$]. (in $\text{ nm}$)

The work function of caesium is $2.14\, eV$. Find the wavelength of the incident light if the photo current is brought to zero by a stopping potential of $0.60\, V$. (Result in $nm$)

Light of wavelength $488 \;nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38 \;V$. Find the work function (in $eV$) of the material from which the emitter is made.

The work function of a metal is $1.6 \ eV$. What is the maximum wavelength of light in $\mathring{A}$ that can cause photoelectric emission from this metal? $(h = 6.6 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

If the frequency of incident radiation $(v)$ is increased,keeping other factors constant,the stopping potential ($v > v_0$,threshold frequency) will: