A photon and an electron have the same energy \( E \). If \( \lambda_p \) is the wavelength of the photon and \( \lambda_e \) is the de Broglie wavelength of the electron, then the ratio \( \frac{\lambda_p}{\lambda_e} \) is:

Question

The de-Broglie wavelength of a moving bus with speed \( v \) is \( \lambda \). Some passengers left the bus at a stop. Now, when the bus moves with twice of its initial speed, its kinetic energy is found to be twice of its initial value. What is the de-Broglie wavelength of the bus now?

Answer 1

The energy of a photon is related to its wavelength by the equation: \[ E = \frac{hc}{\lambda_p} \] Where: - \( h \) represents Planck's constant. - \( c \) is the speed of light. - \( \lambda_p \) denotes the photon's wavelength. For an electron, the de Broglie wavelength is expressed as: \[ \lambda_e = \frac{h}{\sqrt{2mE}} \] Where: - \( m \) signifies the mass of the electron. - \( E \) is the electron's energy. The ratio of these wavelengths is calculated as follows: \[ \frac{\lambda_p}{\lambda_e} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}} = \frac{c}{\sqrt{2mE}} \] Consequently, the wavelength ratio is: \[ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2mE}}{c} \]

A photon and an electron have the same energy \( E \). If \( \lambda_p \) is the wavelength of the photon and \( \lambda_e \) is the de Broglie wavelength of the electron, then the ratio \( \frac{\lambda_p}{\lambda_e} \) is:

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The Correct Option is C

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