Answer
(a) $4.24\times 10^{-7}s$
(b) remains the same
(c) $4.24\times 10^{-7}s$
Work Step by Step
(a) We can find the required time as follows:
$t=\frac{T}{2}$
$t=\frac{1}{2}(\frac{2\pi r}{v})$
$t=\frac{\pi r}{v}$
As $r=\frac{mv}{qB}$
$\implies t=\frac{\pi}{v}(\frac{mv}{qB})=\frac{\pi m}{qB}$
It is also given that the $\alpha$- particle has two protons and two neutrons
$\implies m=2m_p+2m_n$ and $q=2e$
Thus $t=\frac{\pi(2m_p+2m_n)}{(2e)B}$
$\implies t=\frac{\pi(m_p+m_n)}{eB}$
We plug in the known values to obtain:
$t=\frac{3.14(1.67\times 10^{-27}Kg+1.67\times 10^{-27}Kg)}{(1.602\times 10^{-19}C)(0.155T)}$
$t=4.24\times 10^{-7}s$
(b) The equation $t=\frac{\pi(m_p+m_n)}{eB}$ shows that the required time does not depend on the speed of the particle. Hence, the answer to part(a) remains the same even if the speed is doubled.
(c) We know that the time does not depend on the speed, so even if the speed is $2.6\times 10^5m/s$, we obtain the same time, that is $t=4.24\times 10^{-7}s$
