Question

Give a ionisation potential hydrogen atom 13.6 electron volts it exposed electromagetic radiation wavelength 1028 angstroms
give out induced radiation,then

Answer 1

Answer:

Induced radioactivity is a phenomenon which occurs when a stable material has been made radioactive by exposure to specific electromagnetic radiation. Such as in the present case, the H-atom is made radioactive by exposing it to an electromagnetic radiation of wavelength 1028 Angstrom.

Energy of the photon corresponding to wavelength 1028 Angstrom:

Now, As we know that, 1 Joule = 6.24 x 1018 eV, then 1.93 x 10-18 Joule = (1.93 x 10-18 ) x (6.24 x 1018 ) = 12.06 eV

Since, H-atom in its ground state has the energy of 13.6 eV, then:

The energy required to take the electron in the ground state to first excited state is: E1 - E2 = 13.6 - (13.6 / n2) = 13.6 - (13.6 / 22) = 13.6 - 3.4 = 10.2 eV

Similarly, the energy required to take the electron in the ground state to second excited state is: E1 - E3 = 13.6 - (13.6 / 32) = 13.6 - 1.51 = 12.08 eV

Hence, after excitation by the photon of wavelength 1028 Angstrom or the energy of 12.06 eV, it is most probably excited to third stationary state, n = 3. Therefore, the induced radiation from this state will be:

λ1 = n = 3 ----> n = 1 ; λ2 = n = 3 ----> n = 2 ; λ3 = n = 2 ----> n = 1

λ1=hc∆E=hcE1−E36.626×10−34 J s×3×108 ms−1(13.6 − 1.51)×1.6×10−19 J=1.027×10−7 m = 1027 Aoλ2=hc∆E=hcE2−E36.626×10−34 J s×3×108 ms−1(3.4 − 1.51)×1.6×10−19 J=6573 Aoλ3=hc∆E=hcE1−E26.626×10−34 J s×3×108 ms−1(13.6 − 3.4)×1.6×10−19 J=1218 Ao