for its escape (13.6 eV) from the a...
Question  # 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.

JEE/Engineering Exams
Chemistry
Solution 130 4.0 (1 ratings)  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. Quick and Stepwise Solutions Just click and Send OVER 20 LAKH QUESTIONS ANSWERED Download App for Free