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  • The wavelength of electrons accelerated from rest through a potential difference of is <img alt="\mathrm{x} \times 10^{-12} \mathrm{~m}" src="/latex-image/?%5Cmathrm%7Bx%7D%20%5Ctimes

The wavelength of electrons accelerated from rest through a potential difference of 40 \mathrm{kV} is \mathrm{x} \times 10^{-12} \mathrm{~m}. The value of \mathrm{x} is _________(Nearest integer) Given :      Mass of electron =9.1 \times 10^{-31} \mathrm{~kg}
        Charge on an electron =1.6 \times 10^{-19} \mathrm{C}
                Planck's constant =6.63 \times 10^{-34} \mathrm{Js}
 

Answers (1)

best_answer

As we have learnt,

A charge Q coulombs when accelerated by a potential difference of V volts acquires a kinetic energy equal to QV Joules.

\begin{aligned} \therefore K E=Q \times V &=1.6 \times 10^{-19} \times 40 \times 10^{3} \\ &=6.4 \times 10^{-15} \mathrm{~J} . \\ \therefore \quad \text { momentum }(P) &=\sqrt{2 \mathrm{~m} \; K E} \\ &=\sqrt{2 \times 9.1 \times 10^{-31} \times 6.4 \times 10^{-15}} \end{aligned}

                                              = 1.08 \times 10^{-22}

\begin{aligned} \therefore \text { De Broglie's wavelength } &=\frac{h}{P} \\ &=\frac{6.63 \times 10^{-34}}{1.08 \times 10^{-22}} \\ &=6.1 \times 10^{-12} \mathrm{~m} . \end{aligned}

Hence, the correct answer is (6).

Note : Shortcut formula

For an electron accelerated by a potential difference of  V volts, the de Broglie's wavelength can be approximately calculated as 

\lambda \cong \frac{12.25 \AA}{\sqrt{V}}

Posted by

sudhir kumar

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