Chapter 11: Dual Nature of Radiation and Matter (Modern Physics)

An electron accelerated through 30 kV has energy $eV=1.6\times 10^{-19}\times 3\times 10^4=4.8\times 10^{-15}$ J. The most energetic X-ray photon gets all of it.

(a) ${\nu }_{\max }=\frac{eV}{h}=\frac{4.8\times 10^{-15}}{6.63\times 10^{-34}}\approx 7.24\times 10^{18}$ Hz.

(b) ${\lambda }_{\min }=\frac{c}{{\nu }_{\max }}=\frac{3\times 10^8}{7.24\times 10^{18}}\approx 4.14\times 10^{-11}$ m $=0.0414$ nm.

(a) $7.24\times 10^{18}$ Hz. (b) 0.0414 nm.

Photon energy: $h\nu =6.63\times 10^{-34}\times 6\times 10^{14}=3.98\times 10^{-19}$ J $=2.49$ eV.

(a) $K_{\max }=h\nu -{\varphi }_0=2.49-2.14=0.35$ eV.

(b) $e{V}_0=K_{\max }$, so $V_0=0.35$ V.

(c) $K_{\max }=0.35\times 1.6\times 10^{-19}=5.6\times 10^{-20}$ J, and

$$v_{\max }=\sqrt{\frac{2K_{\max }}{m}}=\sqrt{\frac{2\times 5.6\times 10^{-20}}{9.1\times 10^{-31}}}\approx 3.5\times 10^5\text{m/s}$$

(a) 0.35 eV. (b) 0.35 V. (c) About $3.5\times 10^5$ m/s.

$K_{\max }=e{V}_0=1.5$ eV $=1.5\times 1.6\times 10^{-19}=2.4\times 10^{-19}$ J.

1.5 eV, i.e. $2.4\times 10^{-19}$ J.

(a)

$$E=\frac{hc}{\lambda }=\frac{6.63\times 10^{-34}\times 3\times 10^8}{632.8\times 10^{-9}}\approx 3.14\times 10^{-19}\text{J}(\approx 1.96\text{eV})$$

$$p=\frac{h}{\lambda }=\frac{6.63\times 10^{-34}}{632.8\times 10^{-9}}\approx 1.05\times 10^{-27}\text{kg m/s}$$

(b) $N=\frac{P}{E}=\frac{9.42\times 10^{-3}}{3.14\times 10^{-19}}\approx 3.0\times 10^{16}$ photons per second.

(c) $v=\frac{p}{m_H}=\frac{1.05\times 10^{-27}}{1.67\times 10^{-27}}\approx 0.63$ m/s.

(a) $3.14\times 10^{-19}$ J, $1.05\times 10^{-27}$ kg m/s. (b) About $3\times 10^{16}$ per second. (c) About 0.63 m/s.

From $e{V}_0=h\nu -{\varphi }_0$, $V_0=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$. The slope of the $V_0$–$\nu$ graph is $\frac{h}{e}$:

$$h=e\times \text{slope}=1.6\times 10^{-19}\times 4.12\times 10^{-15}\approx 6.59\times 10^{-34}\text{J s}$$

$h\approx 6.59\times 10^{-34}$ J s.

$$V_0=\frac{h(\nu -{\nu }_0)}{e}=\frac{6.63\times 10^{-34}\times (8.2-3.3)\times 10^{14}}{1.6\times 10^{-19}}\approx 2.0\text{V}$$

About 2.0 V.

$$E=\frac{hc}{\lambda }=\frac{1240\text{eV nm}}{330\text{nm}}\approx 3.76\text{eV}$$

This is less than the work function (4.2 eV), so no photoelectrons are emitted, however intense the light.

No: the photon energy (3.76 eV) is less than 4.2 eV.

$h{\nu }_0=h\nu -\frac{1}{2}m{v}_{\max }^2$.

• $\frac{1}{2}m{v}_{\max }^2=\frac{1}{2}\times 9.1\times 10^{-31}\times (6.0\times 10^5{)}^2\approx 1.64\times 10^{-19}$ J
• $h\nu =6.63\times 10^{-34}\times 7.21\times 10^{14}\approx 4.78\times 10^{-19}$ J

$${\nu }_0=\frac{4.78\times 10^{-19}-1.64\times 10^{-19}}{6.63\times 10^{-34}}\approx 4.7\times 10^{14}\text{Hz}$$

About $4.7\times 10^{14}$ Hz.

Photon energy: $E=\frac{1240}{488}\approx 2.54$ eV. $K_{\max }=e{V}_0=0.38$ eV.

$${\varphi }_0=E-K_{\max }=2.54-0.38=2.16\text{eV}$$

About 2.16 eV.

$\lambda =\frac{h}{mv}$:

• (a) $\lambda =\frac{6.63\times 10^{-34}}{0.040\times 1000}\approx 1.7\times 10^{-35}$ m
• (b) $\lambda =\frac{6.63\times 10^{-34}}{0.060\times 1.0}\approx 1.1\times 10^{-32}$ m
• (c) $\lambda =\frac{6.63\times 10^{-34}}{1.0\times 10^{-9}\times 2.2}\approx 3.0\times 10^{-25}$ m

These are far too small to detect, which is why everyday objects show no wave behaviour.

(a) $1.7\times 10^{-35}$ m. (b) $1.1\times 10^{-32}$ m. (c) $3.0\times 10^{-25}$ m.

A photon of radiation with wavelength $\lambda$ and frequency $\nu$ has energy $E=h\nu$ and momentum

$$p=\frac{E}{c}=\frac{h\nu }{c}=\frac{h}{\lambda }$$

Its de Broglie wavelength is

$${\lambda }_{\text{dB}}=\frac{h}{p}=\frac{h}{h/\lambda }=\lambda$$

So the de Broglie wavelength of a photon is the same as the wavelength of the radiation.

Since $p=h/\lambda$ for a photon, ${\lambda }_{\text{dB}}=h/p=\lambda$.