In order to determine the value of `E_(0)`, a scienctist shines photons (''light particles '')of various energies at a cloud of atomic hydrogen. Most of the hydrogen atom occupy transition through that cloud : see figure is a graph of part of the scientist's data, showing the idensity of the transmitted light as a function of the photon energy. A hydrogen atom's electron is likely to obsorb a photon only if the photon gives the electron enough energy to knock it into a higher shell.

According to this experiment , what is the approximate value of `E_(0)`?

Options

A. `1.6 xx 10^(-18) J`
B. `2.1 xx 10^(-18) J`
C. `3.2 xx 10^(-18) J`
D. `6.4 xx 10^(-18) J`

Answer

B

Explanation

From figure (b), photon of energy `1.6 xx 10^(-18) J` get observed in large numbers , no lower energy photon gets obsorbed. And according to the passage, substantial absorption occurs only if the photon jumps the ground state electron into a higher shell.
Therefore, `1.6 xx 10^(-18) J` photon knocks a fround state electron `(n = 1)` into the first excited state `(n = 2)`. Hence , the difference in energy between the ground state and the first excite state must be `1.6 xx 10^(-18) J`.
Using `E_(n) = (E_(0))/(n^(2))`
`1.6 xx 10^(-18) = E_(2) - E_(1) = - (E_(0))/(4) - ((-E_(0))/(1)) = (3)/(4) E_(0)`
`:. E_(0) = (4)/(3) (1.6 xx 10^(-18 J)) = 2.1 xx 10^(-18 J)`