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. `1//e^(2)`
C. `e^(2)`
D. 0

Explanation

Because probability at `x = a_(0)` is porportional to `e^(2s//a_(0) = e^(-2))`
`therefore p =|psi|^(2) prop (e^(-s//a)0)^(2)]`
Probability at nucless `= 0`