What will be the De-Broglie wavelength when the kinetic energy of the electron increases by 5 times?
Correct answer: C. 1/√5
- A. √5
- B. 5
- C. 1/√5
- D. 1/5
Explanation
The de Broglie wavelength (λ) of a particle is given by the λ = where h is the Planck's constant (approximately 6.626 x 10^-34 joule-seconds) and p is the momentum of The momentum (p) of a particle can be calculated using the p = sqrt(2 * m * where m is the mass of the particle and K is the Now, let's consider an electron with an initial kinetic energy (K) and calculate its de Broglie wavelength (λ1). Then, we can calculate the de Broglie wavelength (λ2) when the kinetic energy of the electron increases by 5 Let's assume the initial kinetic energy of the electron is K, and the final kinetic energy is Initial de Broglie wavelength (λ1) = h / sqrt(2 * m * Final de Broglie wavelength (λ2) = h / sqrt(2 * m * (To find the ratio of the final wavelength to the initial wavelength, we can divide λ2 by λ2 / λ1 = (h / sqrt(2 * m * (5K))) / (h / sqrt(2 * m * Simplifying this λ2 / λ1 = sqrt(2 * m * K) / sqrt(2 * m * (5K))= sqrt(K) / sqrt(5K)= sqrt(1/5)= 1 / sqrt(Therefore, when the kinetic energy of the electron increases by 5 times, the de Broglie wavelength of the electron will be 1 / sqrt(5) times the initial de Broglie wavelength.
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Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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