Practice free Dual Nature of Matter and Radiation (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
If intensity of incident light is increased, of photo elctrons will increase.
According to the photoelectric effect, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. Higher intensity means more photons hitting the surface, thus more photoelectrons are emitted.
Frequency of photon having energy 66 eV is............ $(h = 6.6 \times 10^{-34} J.s)$
E = hf $ \therefore f = { E \over h } = { 66 \times 1.6 \times 10^{-19} \over 6.6 \times 10^{-34}} $ $ = 16 \times 10^{15} Hz $
Kinetic energy of proton accelerated under p.d. 1 V will be................
The kinetic energy (K.E.) of a proton when it is accelerated through a potential difference (p.d.) of 1 V is calculated using the equation K.E. = qV, where q is the charge of the proton. Since the charge of a proton is approximately 1e (1 electron volt), the kinetic energy will be 1 eV.
Which of the following phenomenon can not be explained by quantum theory of light ?
Polarisation is a phenomenon that can be explained by the wave theory of light, not by the quantum theory of light. The quantum theory explains phenomena like the photoelectric effect, black body radiation, and the Compton effect.
What will be velocity of light of particle having mass double than its rest mass ?
$ m = { m_0 \over \sqrt {1 -{v^2 \over c^2 }} } = 2m_0 $ $ \therefore \sqrt {1 -{v^2 \over c^2 }}= { 1 \over 2 } $ $ {1 -{v^2 \over c^2 }} = {1 \over 4 } $ $ \therefore {v^2 \over c^2 } = { 3 \over 4 } $ $ \therefore { v \over c }= { \sqrt 3 \over 2 }$ $ \therefore v = { \sqrt3 \over 2 } e $
If de-Broglie wavelength of electron is increased by 1 % its momentum................
$ \lambda = { h \over p } $ $ \therefore \lambda \alpha {1 \over p} $ $ \therefore \lambda \alpha p^{-1} $ $ d \lambda \alpha -{p ^ {-2}}dp $
$ d \lambda \alpha -{ 1 \over p ^ {-2}}dp $ $ \therefore { dp \over \lambda } \times 100 = -{ dp \over p^2} \times p \times 100 $ $\therefore { dp \over \lambda } \times 100 = -1\%$ $ decreases by \%$
With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is $ 0.4 A ^ \circ $
$ \therefore = { h \over \sqrt { 2Vem}} $ $ v = { h^2 \over 2me \lambda ^2 } $ $ = { (6.62 \times 10^{-34})^2 \over 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times (0.4 \times 10^{-1} )^2 }$ = 940.5 =941 V
de-Broglie wavelength of atom at T K absolute temperature will be....................
The de-Broglie wavelength of a particle is given by the formula: $$ ext{wavelength} = rac{h}{ ext{momentum}} $$ For an atom at temperature T, the momentum can be related to the thermal energy as $$ p = rac{mv}{ ext{thermal energy}} $$ The thermal energy is given by $$ rac{3kT}{2} $$ Therefore, the de-Broglie wavelength is $$ ext{wavelength} = rac{h}{ ext{momentum}} = rac{h}{ ext{mv}} rac{1}{ ext{thermal energy}} = rac{h}{ ext{mv}} rac{1}{ ext{(3kT/2)}} = rac{h}{ ext{mv}} rac{1}{ ext{(3kT/2)}} = rac{h}{ ext{sqrt}(3mkT)} $$
Uncertainty of momentum of particle is $ 10^{-30} kg ms^{-1}$ so minimum uncertainty in its position is m.
An electron is accelerated between two points having potential 20 V and 40 V, de- Broglic wavelength of electron is..................
$ \lambda = { h \over \sqrt{2meV} } $ v = v_2 - V_1 = 40 -20 = 20v $ = { 6.62 \times 10^{-34} \over \sqrt { 2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 20}}$ $ = 0.274 \times 19^{-9} $ $ = 2.75 A $
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