Question

Beta and proton are accelerated by the same potential , their de-Broglie ( Mass of proton=mass of neutron), their de-Broglie wavelength have the ratio if the mass of an electron is 2.
(a) $1:2$
(b) $1:4$
(c) $1:1$
(d) $1:33$

Answer 1

Hint: Beta particles are the high moving electron particles emitted by the radioactive decay and we can find the ratio of the de-Broglie wavelength of the beta particle and the proton by using the formula as $\lambda =\dfrac{h}{\sqrt{2mqV}}$. Now solve it.

Complete step by step solution:
First of all, let’s discuss what a beta particle and neurons are. Beta particles are the high-speed electron particles which are emitted by the decay of the radioactive nucleus.
On the other hand, protons are the neutral particles which consist of positive charge and are thus, positively charged particles.
Now considering the statement;
We can find the de-Broglie wavelength by using the relation as;
$\lambda =\dfrac{h}{\sqrt{2mqV}}$
Here, m is the mass, q is the charge and V is the potential difference.
Now we know that;
Both the beta and proton particles are having the same potential i.e. V (given)
Mass of electron is 2m (given)
Mass of the proton is m.
Charge i.e. q on the electron is e.
Charge i.e. q on the proton is e.
Therefore, the ratio of beta particle to proton is;
$\begin{align}
& \dfrac{{{\lambda }_{\beta }}}{{{\lambda }_{p}}}=\dfrac{h}{\sqrt{2mqV}}\times \dfrac{\sqrt{2mqV}}{h} \\
&\implies \dfrac{\sqrt{2meV}}{\sqrt{2\times 2m\times eV}} \\
&\implies \dfrac{1}{\sqrt{2}} \\
&\implies \dfrac{1}{1.414} \\
& \therefore \dfrac{{{\lambda }_{\beta }}}{{{\lambda }_{p}}}=\dfrac{1}{1}=1:1 \\
\end{align}$
Therefore, if the beta and proton are accelerated by the same potential , then their de-Broglie wavelength has the ratio as $1:1$.

Hence, option (c) is correct.

Note: Don’t get confused in the electrons and the beta-particles. Electrons are the negatively charged particles whereas on the other hand, beta particles are emitted only when the radioactive element undergoes nuclear decay and are the fast-moving electrons.