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It work function of Na and Fe are 2.5 eV and 5eV respectively ratio of their threshold frequencies.........................
$ \phi = hf_0 $ $ \therefore \phi \,\alpha f_0 $ $ \therefore {(f_0)_{Na} \over (f_0)_{Fe}} = { \phi _{Na} \over \phi_{Fe} } = { 2.5 \over 5 } = {1 \over 2 } $ = 1:2
If electron is accelerated under the effect of 200V p.d., its kinetic energy = ...................
$ { 1 \over 2 } mv^2 max = eV = 1.6 \times 10^{-19} \times 200 = 3.2 \times 10^{-17} J $
In quantum mechaincs, a particle
In quantum mechanics, a particle can be described as a group of harmonic waves. This concept is fundamental to the wave-particle duality of matter, where particles exhibit both wave-like and particle-like properties. The group of waves forms a wave packet that represents the particle, with the superposition of waves leading to a localized wave packet that corresponds to the particle's position.
The de-Broglie wavelength of a proton and $ \alpha $ - particle is same. The ratio of their velocities will be..............( $ \alpha $ particle is the He-nucleus, having two protonsandtwo neutrons. Thus, its mass $M_\infty = 4 m_p $ where $m_p$ is the mass of the proton.)
$ \lambda_p = \lambda_\alpha $ $ \therefore { h \over m_p v_\alpha } = { h \over m_\alpha v_\alpha } $ $ \therefore { v_p \over v_\alpha} = { m_\alpha \over m_p} $ $ m_\alpha = 4m_p $ $ { v_p \over v_\alpha } = { 4m_p \over m_p} = 4 $ $ \therefore {v_p \over v_\alpha} = 4:1 $
The de-Broglie wavelength associated with a particle with rest mass $m_0$ and moving with speed of light in vacuum is..................
$ \lambda = { h \over mv } $ $ m = { m_0 \over \sqrt {1 - { v^2 \over c^2}} }$ $ \lambda = { h \times \sqrt { 1 - { v^2 \over c^2 } \over m_0 v }} $ $ v =c , \lambda = { h \times \sqrt { 1 - { c^2 \over c^2 } \over m_0 v }} $ $ \therefore \lambda = 0 $
A proton and electron are lying in a box having unpenetrable walls, the ratio of uncertainty in their velocities are........( $m_e $= mass of electron and $m_p $= mass of proton.)
Photons of enerty 1 eV and 2.5 ev successively illuminate a metal, whose work function is 0.5 eV, the ratio of maximum speed of emitted electrion is..........................
$ ( {1 \over 2 mv^2 _{max} }) _1 = (hf)_1 - \phi = 1 - 0.5 $ $ ( {1 \over 2 mv^2_ {max} }) _2 = (hf)_2 - \phi = 2.5 - 0.5 $ $ \therefore {(v^2 _{max} )_1 \over (v^2 _ {max }) _2 } = { 0.5 \over 2 } = { 1 \over 4 } $ $ \therefore {(v^2 _{max} )_1 \over (v^2 _ {max }) _2 } = { 1 \over 2 } = 1:2 $
A proton and an x-particle are passed through same potential difference. If their initial velocity is zero, the ratio of their de Broglie's wavelength after getting accelerated is...................
$ m_p = 1 q_p = e $ $ \lambda = { h \over \sqrt {2meV}} $ $ \therefore \lambda \alpha { 1 \over \sqrt {mq} } $ $ \therefore { \lambda_p \over \lambda_\alpha } = \sqrt { m_\infty q _ \infty \over m_p q _p }$ $ = \sqrt { 4 \times 2e \over 1 \times e } $ $ = \sqrt 8 $ $ = 2 \sqrt 2 $ $ \therefore { \lambda _p \over \lambda_\alpha } = 2 \sqrt {2} : 1 $
The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is.......................
$ \triangle x .\triangle p = { h \over 2 \pi } = h $ $ \triangle x = \lambda $ $ \triangle = {h \over \lambda} $
Wavelength of an electron having energy 10 keV is……….. $ A ^ \circ $
$ \lambda = { h \over \sqrt {2mE} } = { 6.6 \times 10^{-34} \over \sqrt { 2 \times 9.1 \times 10^{-31} \times 10^4 \times 1.6 \times 10^{-19} }}$ $ = 0.12 \times 10^{-10} m $ $ \lambda = 0.12 A ^\circ $
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