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

Solution

E (eV) ≈ 1240/λ (nm).

Solution

$f_0=W_0/h=(2)(1.6\times 10^{-19})/(6.6\times 10^{-34})\approx 4.83\times 10^{14}$ Hz.

Solution

${\text{KE}}_{\max }=h\nu -W_0=4-2=2$ eV.

Solution

$e{V}_0={\text{KE}}_{\max }$. So $V_0=1.5$ V.

Solution

$\lambda =h/p$.

Solution

$\lambda =12.27/\sqrt{V}=12.27/10\approx 1.23$ Å.

Solution

$p=E/c=h\nu /c=h/\lambda$.

Solution

Saturation current ∝ intensity. KE_max stays the same.

Solution

$V_0=(h/e)f-W_0/e$. Slope = h/e.

Solution

${\lambda }_0=1240/W_0(\text{eV})=1240/4=310$ nm.

Solution

Electron diffraction off Ni crystal demonstrates wave nature of matter.

Solution

$V_0=(h\nu -W_0)/e=3$ V.

Solution

$\lambda =h/p$. Equal p → equal λ.

Solution

$\lambda =12.27/\sqrt{400}=12.27/20\approx 0.61$ Å.

Solution

Slope = h is universal; intercepts (f_0) differ → parallel lines.

Solution

$E_{ph}=1240/600=2.07$ eV $=3.3\times 10^{-19}$ J. $N=100/E_{ph}=3\times 10^{20}$/s.

Solution

Photon energy doubles, but W_0 stays. So ${\text{KE}}_{\max }=2h\nu -W_0>2(h\nu -W_0)$. More than doubles.

Solution

Saturation I depends only on intensity, not frequency.

Solution

$\lambda =h/\sqrt{2mqV}$. Ratio = $\sqrt{m_{\alpha }{q}_{\alpha }/(m_p{q}_p)}=\sqrt{4\times 2}=2\sqrt{2}$.

Solution

f > f_0 is the only condition for emission.

Solution

$p=h/\lambda =6.6\times 10^{-34}/(5\times 10^{-7})=1.32\times 10^{-27}$ kg·m/s.

Solution

Low W_0 means visible light gives photoemission.

Solution

$\lambda \propto 1/\sqrt{m}$. ${\lambda }_e/{\lambda }_p=\sqrt{m_p/m_e}=\sqrt{1836}\approx 43$. ${\lambda }_e=43\times 0.4\approx 17$ Å.

Solution

$E_{ph}=1240/200=6.2$ eV. $W_0=E_{ph}-e{V}_0=6.2-1.5=4.7$ eV.

Solution

$E=1240/700\approx 1.77$ eV.

Solution

$f_0=W_0/h=3\times 1.6\times 10^{-19}/(6.6\times 10^{-34})\approx 7.25\times 10^{14}$ Hz.

Solution

λ = 12.27/√100 = 1.23 Å = 0.123 nm.

Solution

Beyond a small V, every electron emitted reaches the collector. Adding more V doesn't increase I.

Solution

$N=I/e=10^{-3}/(1.6\times 10^{-19})=6.25\times 10^{15}$/s.

Solution

$V_0=(h\nu -W_0)/e$. Doubling hν doubles hν but W_0 unchanged → V_0 more than doubles.