Question

The circumference of the second bohr orbit of electron in the h atom is 600nm. calculate the potential diff to which the electron has be subjected so that the electron stops .the electron had the de brogile wavelength , corresponding to the circumference

Answer 1

We know that
2 pi r = n lambda
600 × 10 ^ -6 = 2 lambda
Lambda = 300 * 10^-6
Potential V = h^2/ 2mq(lambda)^2
P = (6.63*10^-34)^2/ 2 *9.1×10^-31 × 1.6×10^-19 ×(600*10^-9)^2
P= 1.675×10^-5 V