NEET Physics Questions: Dual Nature of Matter and Radiation

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In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by





In the Davisson and Germer experiment, the velocity of the electrons emitted from the electron gun can be increased by increasing the potential difference between the anode and the filament. A higher potential difference accelerates the electrons to higher velocities.
A radiation of energy E falls normally on a Perfect reflecting surface. The momentum transferred to the surface is.




Here, surface is perfacet reflector momentum of incident radiation is E/C momentum of reflected rediation is - E/C change in momentum = - Momentum transtered to the surface = $2E \over C$
A photo director area light of wavelength 1400nm Band gap of the Semiconductor used in the photo detector is -----------$ ( h = 6.63 \times 10 ^ {-34} JS ; C = 3 \times 10 ^ 8 m/s )$




No explanation available.
Particle A and B have electric charge + q and + 4 q. Both have mass m. If both are allowed to fall under the same p.d., ratio of velocities vA /vB =




$ W = {1 \over 2} mv^2 = q V $ $ \therefore v = \sqrt {2qV/m} $ $ \therefore {V_A \over V_B} = \sqrt {q_A \over q_B} = \sqrt { q \over 4q} = 1/2 $ $ \therefore { V_A \over V_B } = 1:2 $
Energy of photon having wavelength $ \lambda$ is 2 eV. This photon when incident on metal. maximum velocity of emitted is v. If $\lambda$ is decreased 25% and maximumu velocity is madedouble, work function of metalis ev




$ {1 \over 2 } mv^2_{max} = {hc \over \lambda } - \phi ....(1) $ $ \lambda' = \lambda - 0.25 \lambda = 0.75\lambda \,and \,v' = 2v $ $ \therefore {1 \over 2 } m ( 2v_{max})^2 = { hc \over 0.75 \lambda } -\phi.....(1)$ $4 ( { \lambda c \over \lambda } - \phi) = {4hc \over 3\lambda } - \phi $ $ \therefore {8hc \over 3\lambda} = 3 \phi $ $ \therefore \phi = { 8hc \over 9\lambda } $ $ { hc \over \lambda } = 2eV $ $ \phi = { 8 \over 9 } \times 2eV = 16/9 $ $ \phi = 1.8 eV$
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When elctric bulb having 100 W efficiency emits photon having wavelength 410 mm every second, numbers of photons will be...... $( h =6 \times 10^ {-34} J.s , c = 3 \times 10^8 ms^{-1})$




$ p = { nhc \over \lambda } $ $ \therefore n = { p \lambda \over hc } = { 100 \times 540 \times 10^{-9} \over 6 \times 10^{-34} \times 3 \times 10^8 }$ $ = 3 \times 10^{20} $
de-Broglie wavelength of proton accelerated under 100V electric potential difference is $\lambda _0$ . Ifde - wave length $\alpha$ - particle accelerated by the same electric potential difference will its bouglie




$ {1 \over 2 } mv^2 = eV $ $ \therefore m^2 v^2 = p^2 =2meV$ $ \therefore p = {h \over \sqrt {2meV}}$ $ \therefore p \alpha { 1 \over \sqrt {me}} $ $ \therefore {\lambda_\infty \over \lambda _p } = \sqrt { m_p e_p \over m_\infty e_\infty} $ $ m_\infty = 4 mp ,e_\infty = 2 e_p $ $ \therefore { \lambda _\infty \over \lambda_p} = \sqrt{ m_p e _p \over 4m_p \times 2 e_p } = {1 \over \sqrt 8 } $ $ {\lambda_\infty \over \lambda _p } = { 1 \over 2\sqrt 2} $ $ \therefore {\lambda_\infty \over \lambda_0 } = { 1 \over 2\sqrt 2} $ $ \lambda _8 = {\lambda _0 \over 2\sqrt 2}$
Work function of a body is 4.0 eV. For emission of photoelectron from body, maximum wavelength of light = ..........




$ \phi = 4eV = 4 \times 1.6 \times 10^{-19} J$ $ \therefore \phi = { hc \over \lambda_0} $ $ \therefore \lambda_0 = { hc \over \phi} $ $ ={6.62 \times 10^{-34} \times 3 \times 10^8 \over 4 \times 1.6 \times 10^{-19}}$ $ = 3.103 \times 10^ {-7} m$ $ =310.3 \times 10^ {-9} m$ $ = 310 m $
Photo electric effect on surface is found for frequencies $5.5 \times 10^8 MHz$ and $4.5 \times 10^8 MHz $.If ratio of maximum kinetic energies of emitted photo electrons is 1 : 5, threshold frequency for metal surface is................




$ E = { 1/2} mv^2 max = hf - hf _0 = h (f-f_0) $ $ E_1 = h(f_1-f_0)$ $ E_2 = h (f_2 -f_0)$ $ f_1 = 5.5 \times 10^8 MHz = 5.5 \times 10^{14} Hz $ $ f_2 = 4.5 \times 10^ 8 MHz = 4.5 \times 10^{14} MHz $ $ {E_1 \over E_2} = { f_1 -f_0 \over f_2 - f_0} $ $ { E_1 \over E_2 } = { 1 \over 5 } $ $ \therefore { 1 \over 5 }= { 5.5 \times 10^{14} -f_0 \over 4.5 \times 10^{14} -f_0 } $ $ \therefore 4.5 \times 10^{14} - f_0 = 27.5 \times 10^{14} -5f_0$ $ \therefore 4f_0 = 23.0 \times 10^{14} $ $ \therefore f_0 = { 23 \times 10 ^ { 14} \over 4 }$ $ = 5.75 \times 10^14 Hz$ $ = 5.75 \times 10^8 Hz$
For wave concerned with proton, de-Broglie wavelength change by 0.25% . If its momentum changes by $P_O$ initial momentum = ........




$ \lambda = { \lambda \over p } ...(1) $ $ \therefore \lambda + { 0.25 \over 100} \lambda = { h \over p-p_0} $ $ \lambda {100.25 \lambda \over 100 } = { p \over p-p_0} $ $ \therefore 100.25 p - 100.25 p_0 = 100 p $ $ \therefore 0.25 p =100.25 p_0 $ $ \therefore p = { 100.25 \over 0.25 } $ $ \therefore p = 401 p_0$
According to Einstein's photoelectric equation, graph of kinetic energy of emitted photo electrons from metal versus frequency of incident radation is linear. Its shope




According to Einstein's photoelectric equation, the graph of the kinetic energy of emitted photoelectrons versus the frequency of incident radiation is linear, and the slope of this graph is given by Planck's constant $h$. This slope is the same for all metals and is independent of the intensity of the radiation. Therefore, the correct answer is that the slope is the same for all metals and is free from the intensity of radiation.

Photocell cell is enlightended by small bright source 1 m away. If the same light source is placed 1/2 m away, number of electrons emitted bycathode will be............





I is inversely proportional to 1/r^2

If kinetic energy of free electron is made double, change in de-Broglie wavelength will be............




According to de-Broglie's hypothesis, the wavelength λ is inversely proportional to the momentum p of the particle: $$ ext{λ} = rac{h}{p}$$ where h is Planck's constant. For a free electron, the kinetic energy (K.E.) is given by: $$ ext{K.E.} = rac{p^2}{2m}$$ where m is the mass of the electron. If the kinetic energy is doubled, $$ ext{K.E.}' = 2 imes rac{p^2}{2m} ightarrow ext{p}' = ext{p} imes rac{1}{ ext{√}2}$$. Hence, the new de-Broglie wavelength λ' will be: $$ ext{λ}' = rac{h}{ ext{p}'} = rac{h}{ ext{p} imes rac{1}{ ext{√}2}} = ext{λ} imes ext{√}2$$. Thus, the change in de-Broglie wavelength will be inversely proportional to √2, making the correct answer $$ rac{1}{ ext{√}2}$$.
Energy corresponding to threshold frequency of metatlis 6.2 eV. If stopping potential corresponding to radiation incident on surface is 5V, incident radiation will be in the …….region.




The energy of the incident radiation can be determined using the photoelectric effect equation: $$ ext{Energy of incident radiation} = ext{Threshold energy} + ext{Kinetic energy of ejected electrons}$$. Given the threshold energy is 6.2 eV and the stopping potential is 5V, the total energy of the incident radiation will be: $$6.2 ext{ eV} + 5 ext{ eV} = 11.2 ext{ eV}$$. The corresponding wavelength for this energy falls within the ultraviolet (UV) region of the electromagnetic spectrum.
At $ 10 ^ \circ C $ temperature, de-Broglie wave lengthof atomis $ 0.4 A ^\circ $ . If temperature of atom is increased by $ 30 ^\circ C $ , what will be change in de-Broglie wavelength of atom ?




$ T_1 = 10 +273 = 283 K $ $ T_2 = 40 +273 = 313 K $ $ \lambda _1 = 0.4 A$ $ \lambda = { h \over \sqrt {2mE} }$ $ E = {3 /2 } KT $ $ \lambda = { h \over \sqrt { 3mKT}}$ $ \therefore \lambda \alpha { 1 \over \sqrt T} $ $ \therefore { \lambda_1 \over \lambda_2} = \sqrt { T_1 \over T_2}$ $ \therefore {\lambda_2 \over 0.4 \times 10^{-10}} = \sqrt {283 \over313}$ $ \therefore {\lambda_2 \over 0.4 \times 10^{-10} }= 0.951 $ $ \therefore \lambda_2 = 0.951 \times 0.4 \times 10^{-10}$ $ \therefore \lambda _2 = 0.38 A $ $ = \lambda_2 - \lambda_1 $ $ = 0.38 A ^\circ - 0.4 A ^\circ = - 0.02 A ^\circ $ $ = 2 \times 10^ {-2} A ^\circ decreases $