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An electron of mass me initially at rest moves through a certain distance in a uniform electric field in time t1. A proton of mass mp also initially at rest takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio of \frac{t_2}{t_1}is nearly equal to

  • Option 1)

    1

  • Option 2)

    (\frac{m_p}{m_e})^{\frac{1}{2}}

  • Option 3)

    (\frac{m_e}{m_p})^{\frac{1}{2}}

  • Option 4)

    1836

 

Answers (1)

best_answer

As we have learnt,

 

when Charged Particle at rest in uniform field -

momentum - 

P= mv=m\sqrt{\frac{2Q\Delta v}{m}}= QEt

P=\sqrt{2mQ\Delta v}

-

 

 For electron,

s = \frac{eE}{m_e}\times t_{1}^{2}

for proton,

s = \frac{eE}{m_p}\times t_{2}^{2}

\frac{t_2^{2}}{t_1^{2}} = \frac{m_p}{m_e} \Rightarrow \frac{t_2}{t_1} = \sqrt{\frac{m_p}{m_e}} = (\frac{m_p}{m_e})^{\frac{1}{2}}

 


Option 1)

1

Option 2)

(\frac{m_p}{m_e})^{\frac{1}{2}}

Option 3)

(\frac{m_e}{m_p})^{\frac{1}{2}}

Option 4)

1836

Posted by

Avinash

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