Question

the kinetic energy of a Proton and that of an Alpha particle are four EV and one EV the ratio of de Broglie wavelength associated with them will be

1) 2:1
2) 1:1
3) 1:2
4) 4:1​

Answer 1

Answer:

b) is right answer but you can search it on internet....

Answer 2

The ratio of de Broglie wavelength is 1:1.

(2) is correct option.

Explanation:

Given that,

Kinetic energy of proton = 4 eV

Kinetic energy of alpha particle = 1 eV

We know that,

The de Broglie wavelength is

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

We need to calculate the ratio of de Broglie wavelength

Using formula of de Broglie wavelength

$\dfrac{\lambda_{p}^2}{\lambda_{\alpha}^2}=\dfrac{\dfrac{h^2}{m_{p}^2v_{p}^2}}{\dfrac{h^2}{m_{\alpha}^2v_{\alpha}^2}}$

$\dfrac{\lambda_{p}^2}{\lambda_{\alpha}^{2}}=\dfrac{m_{\alpha}(m_{\alpha}v_{\alpha}^2)}{m_{p}(m_{p}v_{p}^2)}$

Put the value into the formula

$\dfrac{\lambda_{p}^2}{\lambda_{\alpha}^{2}}=\dfrac{4\times m_{p}\times1}{m_{p}\times4}$

$\dfrac{\lambda_{p}^2}{\lambda_{\alpha}^{2}}=\dfrac{1}{1}$

Hence, The ratio of de Broglie wavelength is 1:1.

Learn more :

Topic : de Broglie wavelength

https://brainly.in/question/7584824