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Wave Optics

Question

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The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10^-34 Js, m_e =9.1×10^-31 kg)

5.47A^o

10.9A^o

2.7A^o

None of these

Solution

5.47A^o

$Here : - E = 5 ev = 5 \times 1.6 \times 10^{- 19} J λ = \frac{h}{\sqrt{2 mE}} = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 5 \times 1.6 \times 10^{- 19}}} = \frac{6.6 \times 10^{- 34}}{1.2066 \times 10^{- 24}} = 5.47 \times 10^{- 10} m = 5.47 A^{o}$

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Wave Optics

The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10^-34 Js, m_e =9.1×10^-31 kg)

5.47A^o

10.9A^o

2.7A^o

None of these

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For Daily Free Study Material Join wiredfaculty

Wave Optics

The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10-34 Js, me =9.1×10-31 kg) 5.47Ao 10.9Ao 2.7Ao None of these

Some More Questions From Wave Optics Chapter

Light travels faster in air than that in glass .This is accordance with

In forward biased if the p-n junctions diode is forward biased, then width of potential barrier in p-n junction diode

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The kinetic energy of an electron is 5eV. Calculate the de-Broglie wavelength associated with it (h=6.6×10^-34 Js, m_e =9.1×10^-31 kg)

5.47A^o

10.9A^o

2.7A^o

None of these