The de-Broglie wavelength of an electron (mass 1×10−30kg, charge=1.6×10−19C) with kinetic energy of 200eV is: (Planck's constant 6.6×10−34J):
9.60×10−11m
8.25×10−11m
6.25×10−11m
5.00×10−11m
A
9.60×10−11m
B
6.25×10−11m
C
5.00×10−11m
D
8.25×10−11m
Open in App
Solution
Verified by Toppr
The de-Broglie wavelength λ=h√2mk Given h=6.6×10−34Js m=1×10−30kg K=200eV=200×1.6×10−19J Substituting all these values λ=6.6×10−34√2×1×10−30×200×1.6×10−19 =0.825×10−10=8.25×10−11m
Was this answer helpful?
1
Similar Questions
Q1
The de-Broglie wavelength of an electron (mass 1×10−30kg, charge=1.6×10−19C) with kinetic energy of 200eV is: (Planck's constant 6.6×10−34J):
View Solution
Q2
The de Broglie wavelength of an electron (mass = 1×10−30 kg, charge = 1.6×10−1C) with a kinetic energy of 200 eV is (Planck's constant = 6.6×10−34Js)
View Solution
Q3
The de Broglie wavelength of an electron having 320 eV of energy is nearly (1 eV=1.6×10−19 J, mass of electron =9.1×10−31 kg, Planck's constant =6.6×10−34 Js)
View Solution
Q4
The de-Broglie wavelength of an electron having 80eV of energy is nearly (1eV=1.6×10−19J, mass of electron =9×10−31kg, Planck's constant =6.6×10−34J−s)
View Solution
Q5
An electron is in an excited state in a hydrogen like atom. It has a total energy of –3.4 eV. Find out the value of kinetic energy of the electron (E) and its de-Broglie wavelength λ.