The de Broglie wavelength associated with an electron, accelerated by a potential difference of $$81$$ V is given by:
1. $$13.6$$ nm
2. $$136$$ nm
3. $$1.36$$ nm
4. $$0.136$$ nm
Subtopic: Â De-broglie Wavelength |
Â 54%
From NCERT
NEET - 2023
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The graph which shows the variation of de Broglie wavelength $$(\lambda)$$ of a particle and its associated momentum $$(p)$$ is:

 1 2 3 4
Subtopic: Â De-broglie Wavelength |
Â 81%
From NCERT
NEET - 2022
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NEET 2023 - Target Batch - Aryan Raj Singh

The de-Broglie wavelength of the thermal electron at $$27^\circ \text{C}$$ is $$\lambda.$$ When the temperature is increased to $$927^\circ \text{C},$$ its de-Broglie wavelength will become:
1. $$2\lambda$$
2. $$4\lambda$$
3. $$\frac\lambda2$$
4. $$\frac\lambda4$$

Subtopic: Â De-broglie Wavelength |
Â 60%
From NCERT
NEET - 2022
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An electromagnetic wave of wavelength $$\lambda$$ is incident on a photosensitive surface of negligible work function. If '$$m$$' is the mass of photoelectron emitted from the surface and $$\lambda_d$$ is the de-Broglie wavelength, then:
1. $$\lambda=\left(\frac{2 {mc}}{{h}}\right) \lambda_{{d}}^2$$
2. $$\lambda=\left(\frac{2 {h}}{{mc}}\right) \lambda_{{d}}^2$$
3. $$\lambda=\left(\frac{2 {m}}{{hc}}\right) \lambda_{{d}}^2$$
4. $$\lambda_{{d}}=\left(\frac{2 {mc}}{{h}}\right) \lambda^2$$

Subtopic: Â De-broglie Wavelength |
Â 54%
From NCERT
NEET - 2021
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An electron is accelerated from rest through a potential difference of $$V$$ volt. If the de Broglie wavelength of an electron is $$1.227\times10^{-2}~\text{nm}$$what will be its potential difference?
1. $$10^{2}~\text{V}$$
2. $$10^{3}~\text{V}$$
3. $$10^{4}~\text{V}$$
4. $$10^{5}~\text{V}$$

Subtopic: Â De-broglie Wavelength |
Â 58%
From NCERT
NEET - 2020
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An electron with $$144~\text{eV}$$ of kinetic energy has a de-Broglie wavelength that is very similar to?
1. $$102\times10^{-3}~\text{nm}$$
2. $$102\times10^{-4}~\text{nm}$$
3. $$102\times10^{-5}~\text{nm}$$
4. $$102\times10^{-2}~\text{nm}$$

Subtopic: Â De-broglie Wavelength |
Â 52%
From NCERT
NEET - 2020
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An electron is accelerated through a potential difference of $$10,000~\text{V}$$. Its de-Broglie wavelength is, (nearly):$$\left(m_e = 9\times 10^{-31}~\text{kg}\right )$$
1. $$12.2~\text{nm}$$
2. $$12.2\times 10^{-13}~\text{m}$$
3. $$12.2\times 10^{-12}~\text{m}$$
4. $$12.2\times 10^{-14}~\text{m}$$

Subtopic: Â De-broglie Wavelength |
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From NCERT
NEET - 2019
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A proton and an $$\alpha\text{-}$$particle are accelerated from rest to the same energy. The de-Broglie wavelength $$\lambda_p$$ and $$\lambda_\alpha$$${}_{}$ are in the ratio:
1. $$2:1$$
2. $$1:1$$
3. $$\sqrt{2}:1$$
4. $$4:1$$

Subtopic: Â De-broglie Wavelength |
Â 64%
From NCERT
NEET - 2019
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