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Question:

If the deBroglie wavelength of an electron is equal to 10³ times the wavelength of a photon of frequency 6×10¹⁴ Hz, then the speed of the electron is equal to: (Speed of light = 3×10⁸ m/s, Planck's constant = 6.63×10⁻³⁴ J.s, Mass of electron = 9.1×10⁻³¹ kg)

1.75×10⁶ m/s

1.45×10⁶ m/s

1.8×10⁶ m/s

1.1×10⁶ m/s

Solution:

Let λe be the de Broglie wavelength of the electron and λp be the wavelength of the photon.
Given that λe = 10³λp
The wavelength of a photon is given by λp = c/ν, where c is the speed of light and ν is the frequency.
λp = (3×10⁸ m/s) / (6×10¹⁴ Hz) = 5×10⁻⁷ m
Therefore, λe = 10³ × 5×10⁻⁷ m = 5×10⁻⁴ m
The de Broglie wavelength of an electron is given by λe = h/mv, where h is Planck's constant, m is the mass of the electron, and v is the speed of the electron.
We have: 5×10⁻⁴ m = (6.63×10⁻³⁴ J.s) / (9.1×10⁻³¹ kg × v)
v = (6.63×10⁻³⁴ J.s) / (9.1×10⁻³¹ kg × 5×10⁻⁴ m)
v = 1.456×10⁶ m/s
Rounding to two significant figures, the speed of the electron is approximately 1.5×10⁶ m/s. The closest option is 1.45×10⁶ m/s.