Question

A proton, a neutron, an electron and an $\alpha -\text{particle}$ have same energy. Then their de Broglie wavelengths compare as

• A. ${\lambda }_{\alpha }<{\lambda }_p={\lambda }_n<{\lambda }_e$
  
  
• B. ${\lambda }_e={\lambda }_p={\lambda }_n={\lambda }_{\alpha }$
  
• C. ${\lambda }_c<{\lambda }_p={\lambda }_n>{\lambda }_{\alpha }$
  
  
• D. ${\lambda }_p={\lambda }_n>{\lambda }_e>{\lambda }_{\alpha }$
  
  

Answer 1

The correct option is A ${\lambda }_{\alpha }<{\lambda }_p={\lambda }_n<{\lambda }_e$

According to de-Broglie, the wavelength associated with a particle is given by,$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

As given the energy of all the particles is same.
And we know, $m_{\alpha }>m_p=m_n>m_e$

So,${\lambda }_{\alpha }<{\lambda }_p={\lambda }_n<{\lambda }_e$

Final Answer: $\left(b\right)$