What will be the de – Broglie wavelength when the kinetic energy of the electron increases by 5 times?(a) √5(b) 5(c) \(\frac {1}{\sqrt {5}}\)(d) \(\frac {1}{5}\)I have been asked this question in an interview for internship.I want to ask this question from Wave Nature of Matter topic in portion Dual Nature of Radiation and Matter of Physics – Class 12

What will be the de – Broglie wavelength when the kinetic energy of

1.	What will be the de – Broglie wavelength when the kinetic energy of the electron increases by 5 times?(a) √5(b) 5(c) \(\frac {1}{\sqrt {5}}\)(d) \(\frac {1}{5}\)I have been asked this question in an interview for internship.I want to ask this question from Wave Nature of Matter topic in portion Dual Nature of Radiation and Matter of Physics – Class 12
Answer» Correct answer is (c) \(\frac {1}{\sqrt {5}}\) To explain: The required equation ➔ λ=\(\frac {H}{mv} = \frac {h}{\sqrt {2mK}} \) Where h is the Planck’s constant, m is the mass of the electron and K is the KINETIC energy of the electron. Since the mass of the electron remain unchanged, the wavelength will be inversely proportioned to the kinetic energy, so, \(\frac {\lambda }{\lambda^{‘}} = \sqrt {\frac {K^{‘}}{K} } = \sqrt { \frac {5K}{K} } \) = √5 Therefore, λ’=\(\frac {\lambda }{\sqrt {5}}\) HENCE, the wavelength is reduced by a FACTOR of √5.

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