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The work function of metal is 5.3 eV. What is threshold frequency ?
$ hf_0 = \phi $ $ \therefore f_0 = { \phi \over h } = { 5.3 \times 10^{-19} \times 1.6 \over 6.62 \times 10^ {-34} } Hz $ $ = 1.3 \times 10^{15 } Hz $
An electron moving with velocity 0.6c, then de-brogly wavelength associated with is........... (rest mars of electron, $ m_0 = 9.1 \times 10^{-31} k/s h = 6.63 \times 10^ {-34} Js $)
$ eV_0 = { 1 \over 2 } mv^2 max = 3.8 eV $ $ \therefore V_0 = 3.8 V $ $ {1 \over 2 } mv^2 max = eV_0 = hf - \phi = hf - hf_0 = h ({c \over \lambda} - {c \over \lambda_0} ) $
In an experiment to determine photoelectric charactheristics for a metal the intensity of radiation is kept constant. Strating with threshold frequency. Now, frequency of incident radiation is increased. It is observed that ........
When the frequency of the incident radiation is increased, while keeping the intensity constant, the energy of each photon increases according to the relation $E = h\nu$ (where $\nu$ is the frequency and $h$ is Planck's constant). Therefore, the energy of the emitted photoelectrons will increase.
An oscillator in the walls of cavity in which electromagnetic radiation, has energy equal to 5 hf. Then the oscillator is equivalent to
In the context of electromagnetic radiation in a cavity, the energy of an oscillator being equal to $5hf$ indicates that the oscillator has energy levels that can be occupied by photons. Since the energy is an integer multiple of $hf$, it suggests that the oscillator has a one-to-one correspondence with the photons. Thus, the oscillator is equivalent to 1:1.
Valance electrons in metals
In metals, valence electrons can move freely within the metal but they are not completely free. They move according to their wave functions inside the metal. This is due to the quantum mechanical nature of electrons which describes their behavior in terms of wave functions.
e/m of electrons $ 1.76 \times 10^{11} C / Kg $ and the stopping potential is 0.71 V, themthe maximumvelocity of photo electrons is ........
Radius of a nucleus $ 2 \times 10^{-15} $ . If we imagine an electron inside the nucleus then energy of electron will be = ………….MeV. $m _e = 9.1 \times 10^{-31} kg , h = 6.6 \times 10^ {-34} Js $
$ {4 \pi \times 10^ {20 } }$ $ \triangle x = 2 r = 2 \times 10^{-15} m$ $ \triangle x. \triangle p \approx { h \over 2 \pi } $ $ = { 66 \times 10 ^ {-34} \over 2 \times 2 \times 3.14 \times 2 \times 10h -15 } = 0.5255 \times 10^{-19} $ $ E ={ p^2 \over 2m } P \approx \triangle p $ $ = { (0.5255 \times 10^{-19} )^2 \over 2 \times 9.1 \times 10^{-31} }J = { (0.5255 \times 10^{-19} )^2 \over 2 \times 9.1 \times 10^ {-31} \times 1.6 \times 10^{-19}} $ $ E = 9.48 \times 10^ 3 MeV $
2mW light of wave length $ 4400 A ^ \circ $ is incident on photo sensitive surface of Cs. If quantum efficiency is 0.5 %, what will be the value of photoelectric current ?
$3.56 \mu A$ $ P = { E \over t} $ $ P = {n_1 hc \over \lambda } . t $ $ \therefore n_1 = { p \lambda \over hct } = { 2 \times 10^{-9} \times 44 \times 10 ^ {-8} \over 6.6 \times 10^{-34} \times 3 \times 10^{8} \times 1 } = 4.44 \times 10^9 $ $ n = n_1 of 0.5 \% = 4.4 \times 10^9 \times {0.5 \over 100} $ $ I = ne = 2.22 \times 10h7 \times 1.6 \times 10h -19 = 3.552 $ $ I = 3.56 \times 10^{-6} \mu A$
The difference of kinetic energy of photoelectrons emitted from a surface wavelength $ 2500 A ^ \circ and 5000 A ^ \circ $ will be
$ E_{K1} - E _{K2} = hc ( {1 \over \lambda_1} - { 1 \over \lambda_2 } ) = { hc (\lambda_2 -\lambda_1) \over \lambda_1 \lambda_2 }$ $ \therefore E_{K1} - E_{K2} $ $ = { (6.62 \times 10^{-34}) \times (3 \times 10^8 ) ( 5000 -2500 ) \times 10^{-10} \over (5000 \times 10^{-10} \times (25000) \times 10^{-10}}$ $ = 3.96 \times 10^{-19} J $
Assertion and Reason Type Questions : Assertion : Stopping potential is a measure of K.E. of photo-electrons. Reason : $ W = eV_s = { 1 \over 2 } mv^2 = K.E $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.