For an electron in a hydrogen atom , the wave function `Phi` is proparitional to exp -`r//a_(p)` where `a_(0)` is the Bohr's radius What is the radio of the probability of finding the electron at the nucless at the nucless to the probability of finding id=f at `a_(p)` ?

Options

A. ` e`
B. `e^2`
C. `1//e^2`
D. Zero

Answer

B

Explanation

` psi = 1/( sqrt pi) [1/a_0]^(3//2) e^(-1//a_0)`
`:. pai^2 = 1/(pi) [1/a_0]^3 e^(2r//a_0)`
at nucleus ` r= 0` and in I orbit ` r= a_0`
`:. Psi_0^2 = 1/( pi) [1/a_0]^3 . E^0 = 1/(pi) [1/a_0]^3`
at ` r= a_0`
` :. Psi_n^2 = 1/(pi) [ 1/a_0]^3 . e^(-2)`
`:. (psi_n^2 )/(psi_0^2) =e^2`.