Question

Electron, proton, and $HE^{++}$ are moving with the same KE.T hen order of de-Broglie wavelengths are?

Answer 1

ANSWER.

The order of de - Broglie wavelength are

$\sf \to \: \lambda_{e} > \lambda_{p} > \lambda_{he} {}^{ + + }$

option [ D ] is correct answer.

EXPLANATION.

$\sf \to \: electron \: \: and \: \: proton \: \: and \: \: neutron \: moving \: with \: same \: \: ke \\ \\ \sf \to \: m_{e} = mass \: of \: electron \: = 9.1 \times 10 {}^{ - 31} \\ \\ \sf \to \: m_{p} = mass \: of \: proton = 1.6 \times 10 {}^{ - 27} \\ \\ \sf \to \: m_{he} {}^{ + + } = mass \: of \: helium \: = 4 \times 1.6 \times 10 {}^{ - 27}$

$\sf \to \: as \: we \: know \: that \\ \\ \sf \to \: \lambda \: = \frac{h}{mv} \\ \\ \sf \to \: \lambda \: = \frac{h}{ \sqrt{2mk} } \\ \\ \sf \to \: \lambda_{e} = \frac{1}{ \sqrt{ m_{e} } } \\ \\ \sf \to \: \lambda_{p} = \frac{1}{ \sqrt{ m_{p} } } \\ \\ \sf \to \: \lambda_{he} {}^{ + + } = \frac{1}{ \sqrt{ m_{he} {}^{ + + } } }$

$\sf \to \: order \: of \: the \: wavelength \\ \\ \sf \to \: \lambda_{e} \ratio \: \lambda_{p} \ratio \lambda_{he} {}^{ + + } = \frac{1}{ \sqrt{ m_{e} } } \ratio \: \frac{1}{ \sqrt{ m_{p} } } \ratio \: \frac{1}{ \sqrt{ m_{he} {}^{ + + } } } \\ \\ \sf \to \: \frac{1}{ \sqrt{9.1 \times 10 {}^{ - 31} } } \ratio \: \frac{1}{ \sqrt{1.6 \times 10 {}^{ - 27}} } \ratio \: \frac{1}{ \sqrt{4 \times 1.6 \times 10 {}^{ - 27} } } \\ \\ \sf \to \: \lambda_{e} \ratio \: \lambda_{p} \ratio \lambda_{he} {}^{ + + } \approx \: 84 \ratio \: 2 \ratio \: 1 \\ \\ \sf \to \: \lambda_{e} > \lambda_{p} > \lambda_{he} {}^{ + + }$