The de-Brogilc wavelength of an electron in the first Bohr orbit is

1.	The de-Brogilc wavelength of an electron in the first Bohr orbit is
Answer» `LAMBDA =(1.23)/(sqrt(m))` `lambda = (1.23)/(sqrt(h)) m` `lambda = (1.23)/(sqrtV)` nm `lambda = (1.23)/(V)` SOLUTION :`lambda = 1/(sqrt(2xx 1.6 xx 10^(-19) vm))` `lambda = 1/(sqrt(2xx 1.6 xx 10^(-19) xx v xx 9.11 xx 10^(-31))) = (1.23)/(sqrtV) nm`

Discussion