Question

the ratio of the De Broglie wavelength of a Proton and Alpha particle will be 1:2 if their..​

Answer 1

HELLO THERE!

Here's your answer:

First, the formula for de-Broglie wavelength is:

$\lambda = \frac{h}{mv}$

Where, h is Planck's constant!

Now, given that, the ratio of the wavelength of a proton to that of an alpha particle is 1 : 2.

We know, that the mass of an alpha particle is four times that of mass of a proton. So, if the mass of proton (m₁) = m, then mass of alpha particle (m₂) = 4m.

For the ratio,

$\frac{\lambda_{1}}{\lambda_{2}} = \frac{1}{2}\\\\\implies \frac{\frac{h}{m_{1}v_{1}}}{\frac{h}{m_{2}v_{2}}} = \frac{1}{2}\\\\\implies \frac{m_{2}v_{2}}{m_{1}v_{1}} = \frac{1}{2}\\\\\implies \frac{4m \times v_{2}}{m \times v_{1}} = \frac{1}{2}\\\\\implies \frac{v_{2}}{v_{1}} = \frac{1}{8}\\\\\implies \frac{v_{1}}{v_{2}} = 8:1$

So, the correct answer is Option (B).

Thanks!

Answer 2

Answer:

The correct answer is Option B : Velocity are in the ratio $8:1$

Explanation:

We know that the formula for de-Broglie wavelength is: λ $=\frac{h}{mv}$

where h is the Planck's constant.

In the question it is given that the ratio of the wavelengths of the proton to the alpha particle is $1:2$

Now, we know that mass of the alpha particle is $4$ times the mass of the proton.

So, if the mass of the proton ($m_1$) $=m$,

then mass of the alpha particle ($m_2$)$=4m$

Taking the ratio of the two wavelengths,

λ$_1$ : λ$_2$ $=1:2$

$\frac{\frac{h}{m_1v_1} }{\frac{h}{m_2v_2} } =\frac{1}{2}$

$\frac{m_2v_2}{m_1v_1} =\frac{1}{2}$

Substituting the value of $m_1, m_2$ :

$\frac{4m*v_2}{m*v_1}=\frac{1}{2}$

$\frac{v_2}{v_1}=\frac{1}{8}$

Hence, the ratio of $v_1:v_2=8:1$