N/No = (1/2)ⁿ
t = 2T = 2 x 30 = 60 days

\(\lambda=h c / \Phi=2823 \mathrm{~A}\)

\( \lambda / 2\)

\(\mathrm{Ee}=\mathrm{Mv}^2 / \mathrm{r}\)

Alpha particles due to its largest mass

M_o will not emit photo electron , because its work function is more than 4 ev

The energy of light obtained from the bulb is much less than work function of the wodden block. Hence no photon electrons are emitted.

\(\lambda_{p h}=\lambda=\lambda=h / m v\)
K.E of electrons E \(=1 / 2 m v^2=1 / 2 m\left[h / m^2\right] \lambda\)
\( =\mathrm{h}^2 / 2 \mathrm{~m}^2 \lambda \\ \therefore \mathrm{Eph}=\mathrm{Ee}(2 \mathrm{mc} \lambda / \mathrm{h}) \)

\(E=h c / \lambda\)
= 3.3 x 10^-19 J 

\(\theta=90-\phi / 2\)
\(=90-52 / 2=64^{\circ}\)

stopping potential V₀ = K_max/e = 5ev/e = 5V

Alkali metals have too low work functions. Even visible light can eject electrons from them.

\(E=h v,\)
\( =\mathrm{hc} / \lambda \\ \mathrm{E} \infty 1 / \lambda \)

Power = nhw, where n = No.of photons per second
\(2.0\times { 10 }^{ -3 }=n\times 6.6\times { 10 }^{ -34 }\times 6\times { 10 }^{ 14 }\)
\(n=\frac { 2.0\times { 10 }^{ -3 } }{ 6.6\times { 10 }^{ -34 }\times 6\times { 10 }^{ 14 } } \)
\( =0.050\times { 10 }^{ 17 }=5\times { 10 }^{ 15 } \ Photons/second1/2\)

de-Broglie wavelength of accelerating charged particle is given by
\(\lambda =\frac { h }{ \sqrt { 2mqV } } \Rightarrow \lambda \sqrt { V } =\frac { h }{ \sqrt { 2mq } } =constant\)
(i) The slope of the lines represent \(\frac { h }{ \sqrt { 2mq } } .\)
where, h is Planck's constant, q is the charge and m is the mass of charged particle.
(ii) ₁H² and ₁H³ carry same charge (as they have same atomic number).
\(\therefore \ \lambda \sqrt { V } \propto \frac { 1 }{ \sqrt { m } } \)
The lighter mass i.e ₁H²  is represented by line of greater slope i.e A and similarly ₁H³ by line B.

(i) The de-Broglie wavelength of a particle is given as
\(\lambda=\frac{b}{\sqrt{2 m V_0 q}}\)
Since, \(\alpha\)-particle and proton both are accelerated through the same potential V₀.
\(\begin{aligned} \therefore \quad \lambda \propto \frac{1}{\sqrt{m q}} \end{aligned}\)
 \(\begin{aligned} \text { or } \quad \frac{\lambda_a}{\lambda_p}=\sqrt{\frac{m_p q_p}{m_a q_a}} \end{aligned}\)
As, charge on \(\alpha\)-particle = 2 \(\times\)charge on proton
\(q_{\mathrm{\alpha}}=2 q_{p} \Rightarrow \frac{q_{\mathrm{p}}}{q_{\mathrm{\alpha}}}=\frac{1}{2}\)
Mass of \(\alpha\)-particle = 4 \(\times\) Mass of proton
m_a = 4 \(\times\) m_p
\(\Rightarrow \quad \frac{m_p}{m_\alpha}=\frac{1}{4}\)
\(\begin{array}{ll} \therefore & \frac{\lambda_{\mathrm{a}}}{\lambda_p}=\sqrt{\frac{1}{4} \cdot \frac{1}{2}}=\frac{1}{2 \sqrt{2}}\end{array}\)
\(\begin{array}{ll} \Rightarrow & \lambda_p=2 \sqrt{2 \lambda_\alpha} \end{array}\)
i.e. proton has greater de-Broglie wavelength than that \(\alpha\)-paricle.
(ii) \(\mathrm{KE} \propto q\) (for same accelerating potential)
Since, charge on an \(\alpha\)-paricle is more as compared to a proton, so it will have a greater value of KE. Hence, proton will have lesser KE.

The intensities for both the monochromatic radiations are same but their frequencies are different. It represents
(i) the number of electrons ejected in two cases are same because it depends on the number of incident photons.
(ii) As, \(K{ E }_{ max }=hv-{ \phi }_{ 0 }=hc/\lambda-{ \phi }_{ 0 }\) 
[Einstein's photoelectric equation]
\(\therefore\) The KE_max of violet radiation will be more.

In case of \(_{ 3 }{ { X }^{ 7 } }\),
\(\frac { neutron\quad number }{ proton\quad number } =\frac { 7-3 }{ 3 } =1.33\)
In case of \(_{ 3 }{ { Y }^{ 4 } }\quad \)
 \(\frac { neutron\quad number }{ proton\quad number } =\frac { 4-3 }{ 3 } =\frac { 1 }{ 3 } =0.33\)
For stability, this ratio has to be close to one.
Obviously, nucleus \(_{ 3 }{ { X }^{ 7 } }\) is more stable than the nucleus \(_{ 3 }{ { Y }^{ 4 } }\).

Coulomb forces between protons are repulsive and nuclear forces are ordinarily attractive. For nuclei to be stable nuclear forces must dominate the repulsive forces. Therefore, number of neutrons must be greater than the number of protons.

Let v₁ and v₂ be the velocities of the two particles of masses m₁ and m₂ respectively
According to law of conservation of linear momentum \(\ { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } +{ m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } =M\times 0=0 \)
\( { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } =-{ m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } \ or \ \left| { m }_{ 1 }\overset { \rightarrow }{ { v }_{ 1 } } \right| =\left| { m }_{ 2 }\overset { \rightarrow }{ { v }_{ 2 } } \right| \)
\( { \lambda }_{ 1 }=\frac { h }{ { m }_{ 1 }{ v }_{ 1 } } and{ \lambda }_{ 2 }=\frac { h }{ { m }_{ 2 }{ v }_{ 2 } } \)
\(or \ \frac { { \lambda }_{ 1 } }{ { \lambda }_{ 2 } } =\frac { { m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }{ v }_{ 1 } } =1\)

(i) Energy of a photon,
\(E=hv=\frac { hc }{ \lambda } ,i.e.,E\alpha \frac { 1 }{ \lambda }\)
\(\frac { E' }{ E } =\frac { \lambda }{ 2\lambda } =\frac { 1 }{ 2 } or \ E'=\frac { E }{ 2 } \)
Thus the energy of a photon becomes half when the wavelength of radiation is doubled.
(ii) Momentum of photon,
\(p=\frac { hv }{ c } =\frac { h }{ \lambda } i.e.,\ p\alpha \frac { 1 }{ \lambda } \)
\(\\ \frac { p' }{ p } =\frac { \lambda }{ 2\lambda } =\frac { 1 }{ 2 } or\ p'=\frac { p }{ 2 } \)
Thus the momentum of photon becomes half when the wavelength of radiation is doubled.

Max.K.E. of photoelectron, \({ K }_{ max }=hv-{ \phi }_{ 0 }\) As \({ \phi }_{ 1 }>{ \phi }_{ 2 }\) , so the max. K.E. of the emitted photoelectrons will be greater for the photosensitive material having work function \({ \phi }_{ 2 }\) .

As \({ f }_{ 0 }\) is the threshold frequency, so \({ { \phi }_{ 0 }=hf }_{ 0 }\)
Using Einstein's photoelectric equation, we have 
\(\frac { 1 }{ 2 } { mv }_{ 1 }^{ 2 }=h\times 2{ f }_{ 0 }-{ hf }_{ 0 }={ hf }_{ 0 }\)
\(and \ { \frac { 1 }{ 2 } }{ mv }_{ 2 }^{ 2 }=h\times 5{ f }_{ 0 }-{ hf }_{ 0 }=4{ hf }_{ 0 }\)
\(\\ \frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } =\frac { 1 }{ 4 } \ or \ \frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { 1 }{ 2 } \)

The maximum kinetic energy of the emitted photoelectron from a metal surface is given by 
\({ K }_{ max }=\frac { 1 }{ 2 } { mv }_{ max }^{ 2 }=\frac { hc }{ \lambda } -\ \frac { hc }{ { \lambda }_{ 0 } } =hc\left( \frac { { \lambda }_{ 0 }-\lambda }{ \lambda { \lambda }_{ 0 } } \right) \)
(i)When is \(\lambda >{ \lambda }_{ 0 }\)\({ K }_{ max }\) negative is imaginary. Thus photoelectric emission will not occur.
(ii)When \(\lambda <{ \lambda }_{ 0 }\)\({ K }_{ max }\) is positive, \({ v }_{ max }\) is positive.Thus photoelectric emission will take place and K.E. of photoelectron increases as \(\lambda \)decreases.

If v_o is the threshold frequency, then from photoelectric equation, we have
\({ ev }_{ 1 }=h{ v }_{ 1 }-{ \phi }_{ 0 } \ and \ { ev }_{ 2 }={ hv }_{ 2 }-{ \phi }_{ 0 }\)
\(e({ v }_{ 2 }-{ v }_{ 1 })=h({ v }_{ 2 }-{ v }_{ 1 })or \ h=\frac { e({ v }_{ 2 }-{ v }_{ 1 }) }{ ({ v }_{ 2 }-{ v }_{ 1 }) } \)
\(Now,{ ev }_{ 1 }={ hv }_{ 1 }-{ \phi }_{ 0 }={ hv }_{ 1 }-{ hv }_{ 0 }\)
\(or \ { v }_{ 0 }={ v }_{ 1 }-\frac { { ev }_{ 1 } }{ h } ={ v }_{ 1 }-{ ev }_{ 1 }\left[ \frac { { v }_{ 2 }-{ v }_{ 1 } }{ e({ v }_{ 2 }-{ v }_{ 1 }) } \right]\)
\(={ v }_{ 1 }-\frac { { v }_{ 1 }({ v }_{ 2 }-{ v }_{ 1 }) }{ ({ v }_{ 2 }-{ v }_{ 1 }) } =\frac { { v }_{ 1 }{ V }_{ 2 }-{ v }_{ 1 }{ V }_{ 1 }-{ v }_{ 2 }{ V }_{ 1 }+{ v }_{ 1 }{ V }_{ 1 } }{ ({ v }_{ 2 }-{ v }_{ 1 }) } \)

One incident photon can eject one photoelectron from a photosensitive surface. Therefore the no. of photoelectrons ejected per second depends upon the intensity of the incident light. The increase in the frequency of the incident photon but one photon of high energy can not eject more than one photoelectron from a photosensitive surface.

When an electron in hydrogen atom passes from n=4 energy level to n=1 level, max. a number of photons =6, corresponding to transitions \(4\rightarrow 3;3\rightarrow 2;2\rightarrow 1;4\rightarrow 2,3\rightarrow 1,4\rightarrow 1.\)  The minimum number of photons can be one only corresponding to the transition \(4\rightarrow 1.\)

(i) Slope is determined by h and e. (or slop is independent of the metal used)
(ii) Work function of the metal.

For a particle, de Broglie wavelength, \(\lambda\) = h/p
Kinetic energy, K = p²/2m
Then, \(\lambda\) = h / \(\sqrt{2 m K}\)
For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton \(\left(\begin{array}{l} 1 \\ 1 \end{array} \mathrm{H}\right)\) is 1836 times massive than an electron and an \(\alpha\)-particle \(\left(\begin{array}{l} 4 \\ 2 \end{array} \mathrm{He}\right)\) four times that of a proton. Hence, \(\alpha\) – particle has the shortest de Broglie wavelength.