Light having photon energy `hupsilon` is incident on a metallic plate having work function `phi` to eject the electrons. The most energetic enectrons are then allowed to enter in a region of uniform magnetic field B as shown in Fig. The electrons are projected In X-Z plane making an angle `theta` with X-axis and magnetic field is `vecB=B_0hati` along X-axis maximum pitch of the helix described by an electron is found to be p. Take mass of electron as m and charge as q. Based on the above information, answer the following questions:
Q. The correct relation between p and `B_0` is

Options

A. `qpB_0=2picosthetasqrt(2(hv-phi)m)`
B. `qpB_0=2picosthetasqrt((2(hv-phi))/(m))`
C. `pqB_0=2pisqrt(2(hv-phi)m)`
D. `p=(2pim)/(qB_0)xxsqrt(hv-phi)`

Answer

A

Explanation

At any time t the location of electron is shown as P. In two dimensional view of electron YZ-plane, the situation in more clear.
`v=sqrt((2KE)/(m))=sqrt((2(hv-phi))/m)`
`p=vcostheta(2pim)/(qB_0)`
`pqB_0=2picosthetamsqrt((2(hv-phi))/(m))`
`2picostheta=sqrt(2m(hv-phi))`
X-coordinate, `x=vcosthetaxxt`
Y-coordinate, `y=-[R-Rcosomegat]`
Z-coordinate, `z=Rsinomegat`
So, `z=(mvsintheta)/(qB_0)xxsin[(qB_0)/(m)t]`
`=(sqrt(2m(hv-phi))xxsintheta)/(qB_0)xxsin[(qB_0t)/(m)]`
From `x=vcosthetaxxt=(sqrt(2(hv-phi)))/(m)xxcosthetaxxt`
As v increases, slope of x versus t graph (a straight line) increases