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Dual Nature of Radiation and Matter

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2 months ago

Dual Nature of Radiation and Matter Class 12th

A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option:

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Raj Pandey

Contributor-Level 9

Proton electron

Charge : +e -e

Wavelength : $λ$ $λ$

KE : K_P &

...more

Proton electron

Charge : +e -e

Wavelength : $λ$ $λ$

KE : K_P K_e

Momentum : P_P P_e

As we know KE = $\frac{P^{2}}{2 m}$

Also, de-broglie wavelength, $?$ is given by

$λ = \frac{h}{ρ} o r λ = \frac{h}{\sqrt[]{2 m K}}$

Again $λ = \frac{h}{P}$

$?$ P → same for both

$∴ P_{P} = P_{e}$

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2 months ago

Dual Nature of Radiation and Matter Class 12th

A monochromatic neon lamp with wavelength of 670.5nm illuminates a photo-sensitive material which has a stopping voltage of 0.48V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6nm?

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alok kumar singh

Contributor-Level 10

$e V_{S 2} = \frac{h c}{λ_{1}} - ?$

$e V_{S 2} = \frac{h c}{λ_{2}} - ?$

$e V_{S 2} - e V_{S 1} = \frac{h c}{λ_{2}} - \frac{h c}{λ_{1}}$

$\Rightarrow V_{S 2} = 0.48 + 12.43 * 1 0^{- 7} * (\frac{1}{474.6} - \frac{1}{670.5}) * 1 0^{- 9}$

= 1.25 V

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Dual Nature of Radiation and Matter Class 12th

In a photoelectric experiment, increasing the intensity of incident light:

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2 months ago

Dual Nature of Radiation and Matter Class 12th

The de-Broglie wavelength of a particle having kinetic energy E is $λ .$ How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

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alok kumar singh

Contributor-Level 10

$λ' = λ * \frac{75}{100} = \frac{3}{4} λ$

$λ = \frac{h}{p} = \frac{h}{\sqrt[]{2 m E}}$

$\Rightarrow \frac{E'}{E} = {(\frac{λ}{λ'})}^{2} = {(\frac{4}{3})}^{2}$

$\Rightarrow E' = \frac{16}{9} E$

Extra energy required,

$Δ E = E' - E = \frac{7}{9} E$

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2 months ago

Dual Nature of Radiation and Matter Class 12th

In a photoelectric experiment ultraviolet light o wavelength 280nm is used with lithium cathode having work function $?$ = 2.5eV. If the wavelength of incident light is switched to 400nm, find out the change in the stopping potential. (h = 6.63 * 10^-³⁴ Js, c = 3 * 10⁸ ms^-¹)

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Vishal Baghel

Contributor-Level 10

$\frac{h c}{λ_{1}} = W + e V_{1}$

$\frac{h c}{λ_{2}} = W + e V_{2}$

$= 12400 (\frac{1}{2800} - \frac{1}{4000}) [λ_{1} & λ_{2} a r e t a k e n i n A °]$

$\begin{array}{l} = 4.4 - 3.1 \\ = 1.3 V \end{array}$

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2 months ago

Dual Nature of Radiation and Matter Class 12th

Given below are two statements:

Statement I : Davission-Germer experiment establishes, the wave nature of electrons.

Statement II : If electrons have wave nature, they can interfere and show diffraction.

In the light of the above statements choose the correct answer from the option given below:

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2 months ago

Dual Nature of Radiation and Matter Class 12th

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is:

[Given: The threshold frequency of platinum is 1.3 * 10¹⁵s^-¹ and h = 6.6 * 10^-³⁴ J s.]

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Vishal Baghel

Contributor-Level 10

$?_{0} = h υ_{0}$

$= 6.6 * 1 0^{- 34} * 1.3 * 1 0^{15} J$

$= 8.58 * 1 0^{- 19} J$

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2 months ago

Dual Nature of Radiation and Matter Class 12th

A proton, a neutron, an electron and an a-particle have same energy. If $λ_{p}, λ_{n}, λ_{e} a n d λ_{α}$ and the de Broglie's wavelengths of proton, neutron, electron and a particle respectively, then choose the correct relation form the following:

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alok kumar singh

Contributor-Level 10

$λ = \frac{h}{m v} = \frac{h}{\sqrt[]{2 m k . E}}$

$λ α \frac{1}{\sqrt[]{m}}$

$m_{α} > m_{n} > m_{p} > m_{e}$

$λ_{α} < λ_{n} < λ_{p} < λ_{e}$

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Dual Nature of Radiation and Matter Class 12th

An α particle and a proton are accelerated form rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be

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Payal Gupta

Contributor-Level 10

$k_{α} = 2 e V \Rightarrow P_{α} = \sqrt[]{2 K_{α} m_{α}}$

k_P = eV P_P = $\sqrt[]{2 K_{P} m_{P}}$

$∴ \frac{P_{α}}{P_{P}} = \sqrt[]{\frac{K_{α} m_{α}}{K_{P} m_{P}}} = \sqrt[]{\frac{2 * (4 m_{P})}{m_{P}}} \begin{matrix} \sqrt[]{8} : 1 & 2 \sqrt[]{2} : 1 \end{matrix}$

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2 months ago

Dual Nature of Radiation and Matter Class 12th

An α particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $(λ_{α} : λ_{C 12}) i s :$

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Vishal Baghel

Contributor-Level 10

$λ = \frac{h}{m v} = \frac{h}{\sqrt[]{2 M K . E}}$

$λ α \frac{1}{\sqrt[]{m}}$

$\Rightarrow λ_{α} \sqrt[]{m_{α}} = λ_{C_{12}} \sqrt[]{m_{C_{12}}}$

$\Rightarrow \frac{λ_{α}}{λ_{C_{12}}} = \sqrt[]{\frac{m_{C_{12}}}{m_{α}}} = \sqrt[]{\frac{12}{4}} = \sqrt[]{3} : 1$

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