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11. Dual nature of radiation and matter — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics11. Dual nature of radiation and matter1 Mark
MCQ
A proton and a deuteron are both accelerated through the same potential difference and enter in a magnetic field perpendicular to the direction of the field. If the deuteron follows a path of radius $R ,$ assuming the neutron and proton masses are nearly equal, the radius of the proton' s path will be
  • A
    $\sqrt 2\,R$
  • ✓
    $\frac{R}{{\sqrt 2 }}$
  • C
    $\frac{R}{{2 }}$
  • D
    $R$
✓

Answer

Correct option: B.
$\frac{R}{{\sqrt 2 }}$
b
As charge on both proton and deuteron is same i.e. $'e'$

Energy acquired by both, $E=e V$ For Deuteron.

Kinetic energy, $\frac{1}{2} m V^{2}=e V$

[ $V$ is the potential difference]

$v = \sqrt {\frac{{2eV}}{{{m_d}}}} $

But $m_{d}=2 m$

Therefore, $v=\sqrt{\frac{2 e V}{2 m}}=\sqrt{\frac{e V}{m}}$

Radius of path, $R=\frac{m v}{e B}$

Substituting value of $'V'$ we get

$R = \frac{{2m\sqrt {\frac{{ev}}{m}} }}{{eB}}$

$\frac{R}{2}=\frac{m \sqrt{\frac{e v}{m}}}{e B}$  .... $(i)$

For proton :

$\frac{1}{2} m V^{2}=e V$

$v=\sqrt{\frac{2 e V}{m}}$

Radius of path,  $R' = \frac{{mV}}{{eB}} = \frac{{m\sqrt {\frac{{2eV}}{m}} }}{{eB}}$

$R = \sqrt 2  \times \frac{R}{2}$  [From eq. $(i)$ ]

$R' = \frac{R}{{\sqrt 2 }}$

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