JEE PYQ: Atomic Physics Question 60
Question 60 - 2019 (11 Jan Shift 1)
If the deBroglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14}$ Hz, then the speed of electron is equal to:
(Speed of light $= 3 \times 10^8$ m/s, Planck’s constant $= 6.63 \times 10^{-34}$ J.s, Mass of electron $= 9.1 \times 10^{-31}$ kg)
(1) $1.1 \times 10^6$ m/s
(2) $1.7 \times 10^6$ m/s
(3) $1.8 \times 10^6$ m/s
(4) $1.45 \times 10^6$ m/s
Show Answer
Answer: (4)
Solution
de-Broglie wavelength: $\lambda = \frac{h}{mv} = 10^{-3} \times \frac{3 \times 10^8}{6 \times 10^{14}}$. $v = \frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{9.1 \times 10^{-31} \times 3 \times 10^5} = 1.45 \times 10^6$ m/s.
BETA
