Modern Physics 1 Question 52

48. An electron in a hydrogen like atom is in an excited state. It has a total energy of 3.4eV. Calculate

(1996, 3M)

(a) the kinetic energy,

(b) the de-Broglie wavelength of the electron.

Show Answer

Answer:

Correct Answer: 48. (a) 3.4eV (b) 6.63\AA49.6,3

Solution:

  1. (a) Kinetic energy of electron in the orbits of hydrogen and hydrogen like atoms =∣ Total energy

 Kinetic energy =3.4eV

(b) The de-Broglie wavelength is given by

λ=hp=h2Km

Here, K= kinetic energy of electron Substituting the values, we have

λ=(6.6×1034Js)2(3.4×1.6×1019J)(9.1×1031kg)λ=6.63×1010m or λ=6.63\AA



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक