An electron travels with a velocity of $x m s^{- 1}$ . For a proton to have the same de-Broglie wavelength, the velocity will be approximately:

\frac{1840}{x}

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\frac{x}{1840}

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1840 x

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x

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Solution

The correct option is B $\frac{x}{1840}$
We know that:
de-Broglie wavelength $\Rightarrow λ = \frac{h}{p}$
$1840 \times M_{e} = M_{p}$
$\frac{h}{(M V)_{e}} = \frac{h}{(M V)_{p}}$
$\frac{h}{M_{e} \times x} = \frac{h}{1840 \times M_{e} \times V_{p}}$
$V_{p} = \frac{x}{1840}$

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An electron travels with a velocity of x ms−1. For a proton to have the same de-Broglie wavelength, the velocity will be approximately:

The correct option is B x1840We know that: de-Broglie wavelength ⇒λ=hp 1840×Me=Mp h(MV)e=h(MV)p hMe×x=h1840×Me×Vp Vp=x1840

An electron travels with a velocity of $x m s^{- 1}$ . For a proton to have the same de-Broglie wavelength, the velocity will be approximately:

The correct option is B $\frac{x}{1840}$
We know that:
de-Broglie wavelength $\Rightarrow λ = \frac{h}{p}$
$1840 \times M_{e} = M_{p}$
$\frac{h}{(M V)_{e}} = \frac{h}{(M V)_{p}}$
$\frac{h}{M_{e} \times x} = \frac{h}{1840 \times M_{e} \times V_{p}}$
$V_{p} = \frac{x}{1840}$