If beams of electrons and protons having the same kinetic energy normally enter an electric field, which beam will be more deviated?

Solution:

Question Collected from Telegram Group.

Let r₁ be the path radius followed by the electron and r₂ be the radius path followed by proton.

\[\frac{1}{2}{m_e}v_1^2 = K\]

\[{v_1} = \sqrt {\frac{{2K}}{{{m_e}}}} .......(i)\]

\[\frac{1}{2}{m_p}v_2^2 = K\]

\[{v_2} = \sqrt {\frac{{2K}}{{{m_p}}}} .......(ii)\]

We Know,

\[{\rm{Bqv = }}\frac{{m{v^2}}}{r}........(A)\]

Equation A in terms of Equation (i)

\[Be{v_1} = \frac{{{m_e}v_1^2}}{{{r_1}}}........(iii)\]

\[{r_1} = \frac{{{m_e}{v_1}}}{{Be}}\]

\[{r_1} = \frac{{{m_e}{v_1}}}{{Be}}\]

\[{r_1} = \frac{{{m_e}\sqrt {\frac{{2K}}{{{m_e}}}} .}}{{Be}}.........(iii)\]

Similirly,

\[{r_2} = \frac{{{m_p}\sqrt {\frac{{2K}}{{{m_p}}}} .}}{{Be}}.........(iv)\]

Comparing equation (iii) and (iv),

We come to know that the curve size is inversely proportional to mass of electron or proton.

Since mass of proton is greater than mass of proton.

i.e. m_p>m_e

Thus, we can say r₁>r_2.

Thus, the curve of electron is more as compared to proton.