Q.44. The de Broglie wavelength associated with an electron accelerated through a potential differen

Q: What is the correct answer for JEE Main Physics Q44?

The correct answer is A. Sol. The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where p is momentum. For a charged particle accelerated through a potential difference V, its kinetic energy is $KE = qV$. Also, $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE} = \sqrt{2mqV}$. Thus, $\lambda = \frac{h}{\sqrt{2mqV}}$. For an electron: $\lambda_e = \frac{h}{\sqrt{2m_e e V}}$. For a proton: $\lambda_p = \frac{h}{\sqrt{2m_

Q: Q.44. The de Broglie wavelength associated with an electron accelerated through a potential difference V is and the de Broglie wavelength associated with a proton accelerated through the same…

The correct answer is A. Sol. The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where p is momentum. For a charged particle accelerated through a potential difference V, its kinetic energy is $KE = qV$. Also, $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE} = \sqrt{2mqV}$. Thus, $\lambda = \frac{h}{\sqrt{2mqV}}$. For an electron: $\lambda_e = \frac{h}{\sqrt{2m_e e V}}$. For a proton: $\lambda_p = \frac{h}{\sqrt{2m_

Q: What are the options for this JEE Main Physics question?

A. $\sqrt{\frac{m_p}{m_e}}$ · B. $\frac{m_e}{m_p}$ · C. $\left( \frac{m_p}{m_e} \right)^2$ · D. $\sqrt{\frac{m_e}{m_p}}$

Question

Q.44. The de Broglie wavelength associated with an electron accelerated through a potential difference V is  and the de Broglie wavelength associated with a proton accelerated through the same potential difference is . If their corresponding masses are  and , respectively, then the ratio of their de Broglie wavelengths  is

FirstInTest · Accepted Answer

Sol. The de Broglie wavelength is given by $\lambda = \frac{h}{p}$, where p is momentum. For a charged particle accelerated through a potential difference V, its kinetic energy is $KE = qV$. Also, $KE = \frac{p^2}{2m}$, so $p = \sqrt{2mKE} = \sqrt{2mqV}$. Thus, $\lambda = \frac{h}{\sqrt{2mqV}}$. For an electron: $\lambda_e = \frac{h}{\sqrt{2m_e e V}}$. For a proton: $\lambda_p = \frac{h}{\sqrt{2m_p e V}}$. The ratio $\frac{\lambda_e}{\lambda_p} = \frac{h/\sqrt{2m_e e V}}{h/\sqrt{2m_p e V}} = \frac{\sqrt{2m_p e V}}{\sqrt{2m_e e V}} = \sqrt{\frac{m_p}{m_e}}$. Therefore, option (1) is correct.

Answer

$\sqrt{\frac{m_p}{m_e}}$

Answer

$\frac{m_e}{m_p}$

Answer

$\left( \frac{m_p}{m_e} \right)^2$

Answer

$\sqrt{\frac{m_e}{m_p}}$

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