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Standard XII

Physics

Cut-off Wavelength

The minimum w...

Question

The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volts is equal to:-

\frac{e V}{h c}

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\frac{e h}{c V}

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\frac{h c}{e V}

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\frac{h}{c}

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Solution

The correct option is C $\frac{h c}{e V}$
As we know,
$E = e V = h v_{m a x} = \frac{h c}{λ_{m i n}}$
$\Rightarrow λ_{m i n} = \frac{h c}{e V}$

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The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volts is equal to:-

The correct option is C hceVAs we know,E=eV=hvmax=hcλmin⇒λmin=hceV

The correct option is C $\frac{h c}{e V}$
As we know,
$E = e V = h v_{m a x} = \frac{h c}{λ_{m i n}}$
$\Rightarrow λ_{m i n} = \frac{h c}{e V}$