Q: The allowed energy for the particle for a particular value of n is proportional to :

\$ a = \cfrac{n \lambda}{2}\$De broglie wavelength\$ \lambda = \cfrac{h}{mv}\$Subsitute\$ p =\cfrac{nh}{2a}\$\$ E = \cfrac{1}{2}mv^2= \cfrac{p^2}{2m} =\cfrac{n^2h^2}{8a^2m}\$

Q: A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30\$mW\$and the speed of light is \$3\times 10^{8}m/s\$. The final momentum of the object is:

Sol\$. (B)\$\$t=100\times 10^{-9} sec,\ P=30\times 10^{-3}\$Watt, \$C=3\times 10^{8}m/s\$Incident energy \$= Pt\$By debroglie wavelength, \$\lambda=h/p\$and Energy of photon is \$E=hc/\lambda\$From above two equations, \$E=pc\$ Momentum \$p=\dfrac{Pt}{C}=\dfrac{30\times 10^{-3}\times 100\times 10^{-9}}{3\times 10^{8}}=1.0\times 10^{-17}\$ kg ms\$^{-1} \$So, the answer is option (B).

Q: Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II andindicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.

The X-rays contains \$K_{\alpha}, K_{\beta}\$ etc and these are the transitions between two atomic level . So \$A\rightarrow 1\$Also Hydrogen spectrum has quantized energy levels i.e \$E_n=-\dfrac{136.6}{n^2}\$ and it contains so many transitions between two energy levels like-Lyman, Balmer etc. So \$A\rightarrow 3\$The photo electric effect is the emission of electron when photon incident on a metal. In beta minus decay, one neutron decay into a proton, an electron and an anti-neutrino. So, So \$B \rightarrow 2, 4\$ Mosley's law is an empirical law concerning the characteristic X-rays. So \$C\rightarrow 1\$The Einstein's photoelectric equation is \$h\nu=KE_{max}-h\nu_0\$. So \$D \rightarrow 2\$

Q: The graph between \$ 1/\lambda\$ and stopping potential (V) of three metals having work functions \$\phi_{1},\ \phi_{2}\$ and \$\phi_{3}\$ in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) \$\mathrm{i}\mathrm{s}/\mathrm{a}\mathrm{r}\mathrm{e}\$ correct? [Here \$\lambda\$ is the wavelength of the incident ray].

From photo electric effect, \$eV=\dfrac{hc}{\lambda}-\phi\$ where \$V=\$ stopping potential and \$\phi=\$ work function. Slope , \$m=\tan\theta=hc/e\$ (compare above equation with \$y=mx-c\$)If \$V=0 \Rightarrow \dfrac{\phi}{hc}=\dfrac{1}{\lambda}\$Now from given graph,for metal 1 : \$\dfrac{\phi_1}{hc}=0.001\$for metal 2: \$\dfrac{\phi_2}{hc}=0.002\$for metal 3: \$\dfrac{\phi_3}{hc}=0.004\$Thus, \$\phi_1:\phi_2:\phi_3=1:2:4\$Also, \$\phi=\dfrac{hc}{\lambda_0}\$ where \$\lambda_0\$ be the threshold wavelength. For metal 2 : \$\lambda_{02}=1/0.002=500 nm\$For metal 3 : \$\lambda_{02}=1/0.004=250 nm\$As the wavelength of violet colour is around 400 nm so only metal 2 can eject violet colour light but not metal 3. Thus, options A and C is right.

Q: Light of wavelength \$\displaystyle { \lambda }_{ ph }\$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is \$\displaystyle \phi \$ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is \$\displaystyle { \lambda }_{ e }\$, which of the following statement(s) is(are) true?

\${ \lambda }_{ e }=\$ minimum be broglie wavelength.\$\dfrac { hc }{ { \lambda }_{ ph } } =\dfrac { hc }{ { \lambda }_{ e } } +\phi +eV\$\$\Rightarrow\$ \$hc(\dfrac { -1 }{ { \lambda }_{ ph }^{ 2 } } d{ \lambda }_{ ph })=hc(\dfrac { -1 }{ { \lambda }_{ e }^{ 2 } } d{ \lambda }_{ e })\$\$\Rightarrow\$ \$\dfrac { d{ \lambda }_{ ph } }{ d{ \lambda }_{ e } } =\dfrac { { \lambda }_{ ph }^{ 2 } }{ { \lambda }_{ e }^{ 2 } } \$This denies option A, B and C.\$V>>\dfrac { \phi }{ e } \$Energy of electron,\$E=\dfrac { { P }^{ 2 } }{ 2m } \$\$\lambda_e =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mE } } \$\$=\dfrac{h}{\sqrt{2meV}}\$This supports option D.

Question 1

The allowed energy for the particle for a particular value of n is proportional to :

Accepted Answer

$ a = \cfrac{n \lambda}{2}$De broglie wavelength$ \lambda = \cfrac{h}{mv}$Subsitute$ p =\cfrac{nh}{2a}$$ E = \cfrac{1}{2}mv^2= \cfrac{p^2}{2m} =\cfrac{n^2h^2}{8a^2m}$

Question 2

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 $m W$ and the speed of light is $3 \times 1 0^{8} m / s$ . The final momentum of the object is:

Accepted Answer

Sol$. (B)$$t=100\times 10^{-9} sec,\ P=30\times 10^{-3}$Watt, $C=3\times 10^{8}m/s$Incident energy $= Pt$By debroglie wavelength, $\lambda=h/p$and Energy of photon is $E=hc/\lambda$From above two equations, $E=pc$ Momentum $p=\dfrac{Pt}{C}=\dfrac{30\times 10^{-3}\times 100\times 10^{-9}}{3\times 10^{8}}=1.0\times 10^{-17}$ kg ms$^{-1} $So, the answer is option (B).

Question 3

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II andindicate your answer by darkening appropriate bubbles in the 4  4 matrix given in the ORS.

Accepted Answer

The X-rays contains $K_{\alpha}, K_{\beta}$ etc and these are the transitions between two atomic level . So $A\rightarrow 1$Also Hydrogen spectrum has quantized energy levels i.e $E_n=-\dfrac{136.6}{n^2}$ and it contains so many  transitions between two energy levels like-Lyman, Balmer etc.  So $A\rightarrow 3$The photo electric effect is the emission of electron when photon incident on a metal. In beta minus decay, one neutron decay into a proton, an electron and an anti-neutrino.  So, So $B \rightarrow 2, 4$ Mosley's law is an empirical law concerning the characteristic X-rays.  So $C\rightarrow 1$The Einstein's photoelectric equation is $h\nu=KE_{max}-h\nu_0$. So $D \rightarrow 2$

Question 4

The speed of the particle that can take discrete values is proportional to

Accepted Answer

As given in the questionLet the no. of loops in the wave be  $n$Let wavelength be $\lambda$ Debroglie wavelength$ \lambda = \cfrac{h}{mv}$ ------------(1)Also$ n \lambda = 2 \pi a$  --------------(2)Substituting above$ v = \cfrac{nh}{2am}$$v \propto n$So, the answer is option (D).

Question 5

The graph between $ 1/\lambda$ and stopping potential (V) of three metals having work functions $\phi_{1},\ \phi_{2}$ and $\phi_{3}$ in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) $\mathrm{i}\mathrm{s}/\mathrm{a}\mathrm{r}\mathrm{e}$ correct? [Here $\lambda$ is the wavelength of the incident ray].

Accepted Answer

From photo electric effect, $eV=\dfrac{hc}{\lambda}-\phi$  where $V=$ stopping potential and $\phi=$ work function. Slope , $m=\tan\theta=hc/e$   (compare above equation with $y=mx-c$)If $V=0  \Rightarrow \dfrac{\phi}{hc}=\dfrac{1}{\lambda}$Now from given graph,for metal 1 : $\dfrac{\phi_1}{hc}=0.001$for metal 2: $\dfrac{\phi_2}{hc}=0.002$for metal 3: $\dfrac{\phi_3}{hc}=0.004$Thus, $\phi_1:\phi_2:\phi_3=1:2:4$Also, $\phi=\dfrac{hc}{\lambda_0}$  where $\lambda_0$ be the threshold wavelength. For metal 2 : $\lambda_{02}=1/0.002=500 nm$For metal 3 : $\lambda_{02}=1/0.004=250 nm$As the wavelength of violet colour is around 400 nm so only metal 2 can eject violet colour light but not metal 3. Thus, options A and C is right.

Question 6

Light of wavelength $\displaystyle { \lambda  }_{ ph }$ falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is $\displaystyle \phi $ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is $\displaystyle { \lambda  }_{ e }$, which of the following statement(s) is(are) true?

Accepted Answer

${ \lambda  }_{ e }=$ minimum be broglie wavelength.$\dfrac { hc }{ { \lambda  }_{ ph } } =\dfrac { hc }{ { \lambda  }_{ e } } +\phi +eV$$\Rightarrow$ $hc(\dfrac { -1 }{ { \lambda  }_{ ph }^{ 2 } } d{ \lambda  }_{ ph })=hc(\dfrac { -1 }{ { \lambda  }_{ e }^{ 2 } } d{ \lambda  }_{ e })$$\Rightarrow$ $\dfrac { d{ \lambda  }_{ ph } }{ d{ \lambda  }_{ e } } =\dfrac { { \lambda  }_{ ph }^{ 2 } }{ { \lambda  }_{ e }^{ 2 } } $This denies option A, B and C.$V>>\dfrac { \phi  }{ e } $Energy of electron,$E=\dfrac { { P }^{ 2 } }{ 2m } $$\lambda_e =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mE }  } $$=\dfrac{h}{\sqrt{2meV}}$This supports option D.

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The allowed energy for the particle for a particular value of n is proportional to :

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 $m W$ and the speed of light is $3 \times 1 0^{8} m / s$ . The final momentum of the object is:

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and
indicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.

The speed of the particle that can take discrete values is proportional to

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The allowed energy for the particle for a particular value of n is proportional to :

A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30mWand the speed of light is 3×108m/s. The final momentum of the object is:

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and indicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.

The speed of the particle that can take discrete values is proportional to

The graph between 1/λ and stopping potential (V) of three metals having work functions ϕ1​, ϕ2​ and ϕ3​ in an experiment of photo-electric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? [Here λ is the wavelength of the incident ray].

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A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 $m W$ and the speed of light is $3 \times 1 0^{8} m / s$ . The final momentum of the object is:

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II and
indicate your answer by darkening appropriate bubbles in the 4 4 matrix given in the ORS.