Dual Nature of Radiation and Matter — NCERT Solutions

For X-rays produced by electrons accelerated through $V=30\,\text{kV}=3.0\times10^4\,\text{V}$, the maximum photon energy is

$$E_{\max}=eV=(1.6\times10^{-19})(3.0\times10^4)=4.8\times10^{-15}\,\text{J}.$$

(a) Maximum frequency:

$$\nu_{\max}=\frac{E_{\max}}{h}=\frac{4.8\times10^{-15}}{6.63\times10^{-34}}\approx7.25\times10^{18}\,\text{Hz}.$$

(b) Minimum wavelength:

$$\lambda_{\min}=\frac{c}{\nu_{\max}}=\frac{3.0\times10^8}{7.25\times10^{18}}\approx4.14\times10^{-11}\,\text{m}.$$

Use Einstein's photoelectric equation:

$$K_{\max}=h\nu-\phi_0.$$

Given $\nu=6\times10^{14}\,\text{Hz}$,

$$h\nu=(6.63\times10^{-34})(6\times10^{14})=3.978\times10^{-19}\,\text{J}.$$

In eV,

$$h\nu=\frac{3.978\times10^{-19}}{1.602\times10^{-19}}\approx2.48\,\text{eV}.$$

Work function $\phi_0=2.14\,\text{eV}$.

So,

$$K_{\max}=2.48-2.14=0.34\,\text{eV}.$$

Convert to joules:

$$K_{\max}=0.34\times1.602\times10^{-19}\approx5.45\times10^{-20}\,\text{J}.$$

(b) Stopping potential:

$$eV_0=K_{\max} \Rightarrow V_0=0.34\,\text{V}.$$

(c) Maximum speed:

$$K_{\max}=\frac12 mv^2,$$

with $m_e=9.11\times10^{-31}\,\text{kg}$,

$$v=\sqrt{\frac{2K_{\max}}{m}}=\sqrt{\frac{2(5.45\times10^{-20})}{9.11\times10^{-31}}}\approx3.46\times10^5\,\text{m/s}.$$

The stopping potential is $V_0=1.5\,\text{V}$. From

$$K_{\max}=eV_0,$$

the maximum kinetic energy is

$$K_{\max}=1.5\,\text{eV}.$$

In joules,

$$K_{\max}=1.5\times1.602\times10^{-19}=2.40\times10^{-19}\,\text{J}.$$

Given $\lambda=632.8\,\text{nm}=632.8\times10^{-9}\,\text{m}$ and $P=9.42\,\text{mW}=9.42\times10^{-3}\,\text{W}$.

(a) Energy and momentum of each photon

Photon energy:

$$E=\frac{hc}{\lambda}$$

$$E=\frac{(6.63\times10^{-34})(3.0\times10^8)}{632.8\times10^{-9}}\approx3.14\times10^{-19}\,\text{J}.$$

Photon momentum:

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

$$p=\frac{6.63\times10^{-34}}{632.8\times10^{-9}}\approx1.05\times10^{-27}\,\text{kg m/s}.$$

(b) Photons per second

$$N=\frac{P}{E}=\frac{9.42\times10^{-3}}{3.14\times10^{-19}}\approx3.0\times10^{16}\,\text{photons/s}.$$

(c) Speed of hydrogen atom with same momentum

For a hydrogen atom, $p=mv$.

Using $m_H\approx1.67\times10^{-27}\,\text{kg}$,

$$v=\frac{p}{m_H}=\frac{1.05\times10^{-27}}{1.67\times10^{-27}}\approx6.27\times10^{-4}\,\text{m/s}.$$

From Einstein's photoelectric equation,

$$V_0=\frac{h}{e}\nu-\frac{\phi_0}{e},$$

so the slope of the $V_0$ vs $\nu$ graph is

$$\frac{h}{e}.$$

Given slope $=4.12\times10^{-15}\,\text{V s}$,

$$h=e\times\text{slope}=(1.602\times10^{-19})(4.12\times10^{-15})\approx6.60\times10^{-34}\,\text{J s}.$$

Using

$$eV_0=h(\nu-\nu_0),$$

or in eV units,

$$V_0=\frac{h}{e}(\nu-\nu_0).$$

Here,

$$\nu-\nu_0=(8.2-3.3)\times10^{14}=4.9\times10^{14}\,\text{Hz}.$$

Now

$$\frac{h}{e}\approx4.14\times10^{-15}\,\text{V s},$$

so

$$V_0=(4.14\times10^{-15})(4.9\times10^{14})\approx2.03\,\text{V}.$$

But the textbook uses the standard relation with precise values; using $h=6.63\times10^{-34}\,\text{J s}$ and $e=1.6\times10^{-19}\,\text{C}$ gives

$$V_0=\frac{6.63\times10^{-34}}{1.6\times10^{-19}}\times4.9\times10^{14}\approx2.03\,\text{V}.$$

So the predicted cut-off voltage is $2.03\,\text{V}$.