A) \[\sqrt{2}R\]
B) \[\frac{R}{\sqrt{2}}\]
C) \[\frac{R}{2}\]
D) R
Correct Answer: B
Solution :
| [b] As charge on both proton and deuteron is same i.e. 'e' |
| Energy acquired by both, E = eV |
| For Deuteron. |
| Kinetic energy, \[=\frac{1}{2}m{{V}^{2}}=eV\] |
| [V is the potential difference]\[v=\sqrt{\frac{2eV}{{{m}_{d}}}}\] |
| But\[{{m}_{d}}=2m\] |
| Therefore, \[v=\sqrt{\frac{2eV}{2m}}=\sqrt{\frac{eV}{m}}\] |
| Radius of path, \[R=\frac{mv}{eB}\] |
| Substituting value of V we get\[R=\frac{2m\sqrt{\frac{ev}{m}}}{eB}\] |
| \[\frac{R}{2}=\frac{m\sqrt{\frac{ev}{m}}}{eB}\] (i) |
| For proton: |
| \[\frac{1}{2}m{{V}^{2}}=eV\] |
| \[V=\sqrt{\frac{2eV}{m}}\] |
| Radius of path, \[R'=\frac{mV}{eB}=\frac{m\sqrt{\frac{2eV}{m}}}{eB}\] |
| \[R'=\sqrt{2}\times \frac{R}{2}\][From eq. (i)]\[R'=\frac{R}{2}\] |
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