Atomic StructureHard
Question
Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principal quantum number ′n′. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is 12.75 eV. Then find the principal quantum number ′n′ of the excited state.
Options
A.1
B.2
C.3
D.4
Solution
KE1 = Ephoton – BEn = 1
KE2 = Ephoton – BEn = n
KE2 - KE1 = BEn = 1 - BEn = n = 13.6 Z2
= 12.75 (given).
∴ n2 = 16 or n = 4.
BE : Binding energy.
KE2 = Ephoton – BEn = n
KE2 - KE1 = BEn = 1 - BEn = n = 13.6 Z2
= 12.75 (given).∴ n2 = 16 or n = 4.
BE : Binding energy.
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The ratio of the ″e/m″ (specific charge) values of a electron and an a-particle is -...If wavelength is equal to the distance travelled by the electron in one second, then -...An α-particle is accelerated through a potential difference of V volts from rest. The de-Broglie′s wavelength...The mass of a particle is 10–10 g and its diameter is 10−4 cm. If its speed is 10–6 cm/s with 0.0001% uncertainty in mea...The configuration [Ar] 3d10 4s2 4p4 is similar to that of...