When Barium is irradiated by a light of \[\lambda =4000\overset{{o}}{\mathop{{A}}}\,\] all the photoelectrons emitted are bent in a circle of radius 50 cm by a magnetic field to flux density \[5.26\times {{10}^{-6}}T\] acting perpendicular to plane of emission of photoelectrons. Then:
(The question has multiple correct choices)
A) The kinetic energy of the fastest photoelectron is $0.6eV$.
B) Work function of the metal is $2.5 eV$.
C) The maximum velocity of the photoelectron is \[0.46\times {{10}^{6}}m{{s}^{-1}}\].
D) The stopping potential for the photoelectric effect is $0.6eV.$

Hint: To find the kinetic energy and work function, we first need to find the velocity of the photoelectron and to find the velocity of photoelectron we use the formula of:
$v$ = $\dfrac{Bqr}{m}$
After finding the velocity, we use the kinetic energy formula of:
Kinetic Energy (K.E.) \[=\dfrac{1}{2}m{{v}^{2}}\]
And then to find the work function we use the formula of:
$\Phi$ = $\dfrac{hc}{\lambda}$ - $K.E.$
where h is the Planck’s constant, c is the speed of light, $\lambda $ is the wavelength, B is the magnetic flux, q is the charge as \[1.6\times {{10}^{-19}}\]kg, r is the radius to which the magnetic field was bend and m is the mass of electron as \[9.1\times {{10}^{-31}}\].

Complete step by step solution:
As the question says that the question have multiple correct options, do let us first find the velocity of the photoelectron moving by using the formula:
$v$ = $\dfrac{Bqr}{m}$
Now let us place the values from the question in the above formula with the value of the charge q as
and changing the radius dimension from cm to m as
$q$ = $1.6\times {{10}^{-19}}C$ and $50cm=0.5m$
Hence, the velocity of the photoelectron is:
\[\Rightarrow v{ }={ }\dfrac{5.26\times {{10}^{-6}}\times 1.6\times {{10}^{-19}}\times 0.5}{9.1\times {{10}^{-31}}}\]
\[\Rightarrow v{ }=0.46\times {{10}^{6}}m{{s}^{-1}}\]
This matches with one of the options i.e. C hence, correct.
After this we find the kinetic energy as suggested in option A to see if it matches the value of 0.6 eV.
Placing the values of the mass and velocity in the kinetic energy formula we get the value of the kinetic energy as:
\[\Rightarrow \dfrac{1}{2}m{{v}^{2}}\]
\[\Rightarrow \dfrac{1}{2}\times 9.1\times {{10}^{-31}}\times {{(0.46\times {{10}^{6}})}^{2}}\]
\[\Rightarrow K.E=0.6eV\]
Hence, option A is correct as well.

Moving onto the stopping potential the kinetic energy that we found is the maximum kinetic energy needed for the photoelectric effect and when that happens the stopping potential is equal to the kinetic energy of the electron hence, option D is true.
Now after getting the kinetic energy and have the means to calculate binding energy we find the value of work function using the formula of:
\[\Phi =\dfrac{hc}{\lambda }-\] K.E.
\[\Rightarrow \Phi =\dfrac{6.62607004{ }\times { }{{10}^{-34}}~{{m}^{2}}~kg{{s}^{-1}}\times 3\times {{10}^{8}}}{400}-0.6\]
\[\Rightarrow \Phi =2.5eV\]

Therefore, we can say that all the options are correct.

Note: The formula for binding energy is \[\dfrac{hc}{\lambda }\] and the stopping potential means the voltage required to stop electrons between plates to form photoelectric current.