Dual Nature of Radiation and MatterNEET Physics · Class 12 · NCERT Chapter 11

Solution

The kinetic energy of photoelectrons is given by $K=h\nu -\varphi$, where $\varphi$ is the work function. Given $K_1=0.5\text{eV}$ and $K_2=0.8\text{eV}$, and the energy of the incident radiation increases by 20%, we have $h{\nu }_2=1.2h{\nu }_1$. Substituting, $0.8=1.2h{\nu }_1-\varphi$ and $0.5=h{\nu }_1-\varphi$. Solving these, $\varphi =1.0\text{eV}$, so option (b) is correct.

Solution

The de-Broglie wavelength $\lambda$ is inversely proportional to the square root of the kinetic energy $K$, i.e., $\lambda \propto \frac{1}{\sqrt{K}}$. If $K$ is increased by 16 times, $\lambda$ becomes $\frac{\lambda }{4}$. The percentage change in $\lambda$ is $\left(1-\frac{1}{4}\right)\times 100=75\%$, so option (b) is correct.

Solution

KEmax=hc λ−Ψ KEmax=1240 500−2.82 KEmax=2.48−2.28=0.2 eV www.vedantu.com 42 λmin= h √2m(KE)max = 20 3 ×10−34 √2×9× 10−31×0.2×1.6× 10−19 λmin=25 9 ×10−9=2.80×10−9 nm So λ≥2.8× 10−9 m

Solution

𝑘1=ℎ𝑐 𝜆 − 𝜓 𝑘2=3𝑘1=2ℎ𝑐 𝜆 −𝜓= 3ℎ𝑐 𝜆 −3𝜓 So 2 𝜓= ℎ𝑐 𝜆 So 𝜓= ℎ𝑐 2𝜆

Solution

De-Broglie wavelength is given by λe = h p = h √2m.E for electron De-Broglie wavelength of photon is given by λp = h p = h E c = hc E λe λp = 1 √2mE .E c = 1 c√ E 2m

Solution

In photo electric effects eV0 = 48− W eV0 = hc λ − W eV= hc λ − W …(i) e V 4 = hc 2λ − W ….(ii) From (i) and (ii) hc λ − W = 4(hc 2λ− W) hc λ − W =2hc λ − 4W 3W = hc λ ⇒ W =hc 3λ hc λmax = hc 3λ ⇒ λmax = threshold wavelength 3λ www.vedantu.com 20

Solution

de-Broglie wavelength h mvλ= = 2( K E ) h m 32 ()2 h mk T = 3 h mkT λ=

Solution

λ0 = 3250 × 10 –10 m λ = 2536 × 10–10 m φ = 1242 eV-nm 3.82 eV325 nm = hν = 1242 eV-nm 4.89 eV253.6 nm = KEmax = (4.89 – 3.82) eV = 1.077 eV 21 91 1.077 1.6 102 mv -=× × v = 19 31 2 1.077 1.6 10 9.1 10 - - ×× × × v = 0.6 × 10 6 m/s

Solution

Energy of 2 s-orbital and 2 p-orbital in case of hydrogen like atoms is equal.

Solution

Using the photoelectric equation $E_k=h\nu -\varphi$, where $\varphi =h{\nu }_0$. For $\nu =2{\nu }_0$, $E_{k1}=h(2{\nu }_0-{\nu }_0)=h{\nu }_0$. For $\nu =5{\nu }_0$, $E_{k2}=h(5{\nu }_0-{\nu }_0)=4h{\nu }_0$. Since $E_k=\frac{1}{2}m{v}^2$, $\frac{v_1^2}{v_2^2}=\frac{h{\nu }_0}{4h{\nu }_0}=\frac{1}{4}\Rightarrow \frac{v_1}{v_2}=\frac{1}{2}$. Thus, the ratio $v_1:v_2$ is $1:2$, so option (d) is correct.

Solution

To convert joules to electron-volts, use the conversion factor $1\text{eV}=1.6\times 10^{-19}\text{J}$. For $10^{-20}\text{J}$, the energy in eV is $\frac{10^{-20}}{1.6\times 10^{-19}}\approx 0.06\text{eV}$, so option (c) is correct.

Solution

The de Broglie wavelength $\lambda$ of an electron is given by $\lambda =\frac{h}{\sqrt{2meV}}$, where $h$ is Planck's constant, $m$ is the mass of the electron, and $e$ is the charge of the electron. Substituting the given values, $1.227\times 10^{-2}\times 10^{-9}=\frac{6.626\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.602\times 10^{-19}\times V}}\Rightarrow V=10\text{V}$, so option (a) is correct.

Solution

The photoelectric current depends on the number of photoelectrons emitted, which is proportional to the intensity of the incident light. Halving the frequency to below the threshold frequency means no photoelectrons are emitted, but since the intensity is doubled, the number of photons is doubled, leading to a doubled photoelectric current. Thus, option (a) is correct.

Solution

The de-Broglie wavelength of the photoelectron is given by ${\lambda }_d=\frac{h}{p}$, where $p$ is the momentum. The momentum of the photoelectron can be derived from the energy of the incident photon, $E=\frac{hc}{\lambda }=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$. Solving for $p$, we get $p=\sqrt{\frac{2m{c}^2\lambda }{h}}$. Substituting this into the de-Broglie equation, ${\lambda }_d=\frac{h}{\sqrt{\frac{2m{c}^2\lambda }{h}}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h\sqrt{h}}{\sqrt{2m{c}^2\lambda }}=\frac{h^{3/2}}{\sqrt{2m{c}^2\lambda }}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}=\frac{h}{\sqrt{2m{c}^2\lambda /h}}$. Simplifying, $\lambda_d = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^2 \lambda / h}} = \frac{h}{\sqrt{2mc^

Solution

de-Broglie wavelength associated with a particle is given by h pλ= 1 pλ∝

Solution

Since max sk eV h= = ν−φ 02 = ν− νSeV hh …(i) 02 ν= −νS heV h …(ii) 00 1 22 ν −ν =ν −ν h h hh ⇒ 0 0 24 ν νν − = ν−h hhh ⇒ 0 3 24 ν ν=h h 0 3 2 νν= * Language of question is wrongly framed. The values of stopping potentials should be interchanged.

Solution

The minimum wavelength ${\lambda }_{\text{min}}$ of X-rays is given by the equation ${\lambda }_{\text{min}}=\frac{hc}{eV}$, where $h$ is Planck's constant, $c$ is the speed of light, and $e$ is the charge of an electron. This shows that ${\lambda }_{\text{min}}$ is inversely proportional to $V$, so option (c) is correct.

Solution

The incident energy of 2.20 eV is greater than the work function of Caesium (2.14 eV) but less than that of Potassium (2.30 eV) and Sodium (2.75 eV). Therefore, only Caesium can emit photoelectrons, so option (b) is correct.

Solution

de-Broglie wavelength 2 h h h P mv mE λ = = = where 21 2E mv= Squaring both sides, 22 24 h mE λ= 2 1 (constant) E⇒= λ Graph passes through origin with constant slope. - 7 - NEET (UG)-2024 (Code-Q1)

Solution

(A) If c is the velocity of light so, E = hν (Energy of photon) (B) Velocity of photon is equal to velocity of light i.e. c. (C) λ= h p = λ hp ν= hp c (D) In photon-electron collision both total energy and total momentum are conserved.

Solution

For photon, $E=\frac{hc}{{\lambda }_{ph}}\Rightarrow {\lambda }_{ph}=\frac{hc}{E}$

For electron, $p=\text{momentum}$ and $E=\frac{p^2}{2m}={\left(\frac{h}{{\lambda }_e}\right)}^2\times \frac{1}{2m}$

$\Rightarrow {\lambda }_e=\frac{h}{\sqrt{2mE}}$

$\therefore \frac{{\lambda }_{Ph}}{{\lambda }_e}=\frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}}=c\sqrt{\frac{2m}{E}}$

Solution

Photoelectric current is directly proportional to intensity of light.