The de-Broglie wavelength of a particle accelerated with 200 volt potential is 6A0. If it is accelerated by 1800 volts p.d. its wavelength will be:

λ=h2mqV λα1Vλ1λ2=V2V1 6λ2=18002006λ2=3 ⇒λ2=2A0

The de-Broglie wavelength of a particle accelerated with 200 volt potential is 6A0. If it is accelerated by 1800 volts p.d. its wavelength will be:

λ=h2mqV λα1Vλ1λ2=V2V1 6λ2=18002006λ2=3 ⇒λ2=2A0

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The de-Broglie wavelength of a particle accelerated with 200 volt potential is $6 \overset{0}{A}$ . If it is accelerated by 1800 volts p.d. its wavelength will be:

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1 $\overset{0}{A}$

2 $\overset{0}{A}$

3 $\overset{0}{A}$

None

answer is B.

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$λ = \frac{h}{\sqrt{2 m q V}} λ α \frac{1}{\sqrt{V}}$

$\frac{λ_{1}}{λ_{2}} = \sqrt{\frac{V_{2}}{V_{1}}} \frac{6}{λ_{2}} = \sqrt{\frac{1800}{200}}$

$\frac{6}{λ_{2}} = 3 \Rightarrow λ_{2} = 2 \overset{0}{A}$

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The de-Broglie wavelength of a particle accelerated with 200 volt potential is 6A0. If it is accelerated by 1800 volts p.d. its wavelength will be:

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The de-Broglie wavelength of a particle accelerated with 200 volt potential is $6 \overset{0}{A}$ . If it is accelerated by 1800 volts p.d. its wavelength will be: