Question

de-Broglie wavelength for electron ( left(lambda_{mathrm{e}}right)=frac{h}{m_{e} V_{e}} )
Similarly, de-Broglie wavelength for proton ( left(lambda_{mathrm{p}}right)= ) ( frac{h}{m_{p} v_{p}}=frac{h}{1836 m_{e} v_{p}}[sin c e p r o t o n i s 1836 t i m e s h e a v i e r t h a n text { electron }] )
The condition when the de-Broglie wavelengths associated with an electron would be equal to that associated with a proton, i.e. when ( lambda_{e}=lambda_{p} )
( frac{h}{m_{e} V_{e}}=frac{h}{1836 m_{e} V_{p}} )
( frac{1}{V_{e}}=frac{1}{1836 V_{p}} )
or ( V_{e}=1836 V_{p} )
Hence, when velocity of an electron is equal to 1836 times the velocity of proton, the de-Broglie wavelengths associated with an electron would be equal to that associated with a proton.

# for its escape (13.6 eV) from the atom. What is wavelength of the emittede E-2 Deuuce the condition when the De-Broglie wavelenath associated with an electron would be equal to that associated with a proton if a proton is 1836 times heavier than an electron natalia voet is initially accelerated through a potential difference of 100 volts.

Solution