The maximum photon energy is the kinetic energy gained by an electron accelerated through $30\,\text{kV}$: $E=eV=30\,\text{keV}=30000(1.602\times10^{-19})\,\text{J}$. Thus $\nu_{\max}=E/h=(4.806\times10^{-15})/(6.626\times10^{-34})=7.25\times10^{18}\,\text{Hz}$. The minimum wavelength is $\lambda_{\min}=c/\nu_{\max}=hc/eV=4.14\times10^{-11}\,\text{m}$.
(a) $7.25\times10^{18}\,\text{Hz}$. (b) $4.14\times10^{-11}\,\text{m}$, or $0.0414\,\text{nm}$.
Photon energy is $h\nu=(6.626\times10^{-34})(6.0\times10^{14})=3.98\times10^{-19}\,\text{J}=2.48\,\text{eV}$. Hence $K_{\max}=h\nu-\phi=2.48-2.14=0.34\,\text{eV}=5.5\times10^{-20}\,\text{J}$. Since $eV_0=K_{\max}$, $V_0=0.34\,\text{V}$. From $K_{\max}=\frac12mv^2$, $v=\sqrt{2K/m}=\sqrt{2(5.5\times10^{-20})/(9.11\times10^{-31})}=3.5\times10^5\,\text{m s}^{-1}$.
(a) $0.34\,\text{eV}=5.5\times10^{-20}\,\text{J}$. (b) $0.34\,\text{V}$. (c) $3.5\times10^5\,\text{m s}^{-1}$.
The stopping potential satisfies $K_{\max}=eV_0$. For $V_0=1.5\,\text{V}$, $K_{\max}=1.5\,\text{eV}=1.5(1.602\times10^{-19})=2.4\times10^{-19}\,\text{J}$.
$1.5\,\text{eV}=2.4\times10^{-19}\,\text{J}$.
For one photon, $E=hc/\lambda=(6.626\times10^{-34})(3.0\times10^8)/(632.8\times10^{-9})=3.14\times10^{-19}\,\text{J}$. Momentum is $p=h/\lambda=6.626\times10^{-34}/(632.8\times10^{-9})=1.05\times10^{-27}\,\text{kg m s}^{-1}$. The photon rate is $N=P/E=(9.42\times10^{-3})/(3.14\times10^{-19})=3.0\times10^{16}\,\text{s}^{-1}$. For a hydrogen atom, $v=p/m_H=(1.05\times10^{-27})/(1.67\times10^{-27})=0.63\,\text{m s}^{-1}$.
(a) $E=3.14\times10^{-19}\,\text{J}$ and $p=1.05\times10^{-27}\,\text{kg m s}^{-1}$. (b) $3.0\times10^{16}\,\text{photons s}^{-1}$. (c) $0.63\,\text{m s}^{-1}$.
Einstein's equation gives $V_0=(h/e)\nu-\phi/e$, so the slope of the $V_0$ versus $\nu$ graph is $h/e$. Therefore $h=e\times\text{slope}=(1.602\times10^{-19})(4.12\times10^{-15})=6.60\times10^{-34}\,\text{J s}$.
$6.60\times10^{-34}\,\text{J s}$.
For photoelectric emission, $eV_0=h(\nu-\nu_0)$. Hence $V_0=(h/e)(8.2-3.3)\times10^{14}=(4.14\times10^{-15})(4.9\times10^{14})=2.03\,\text{V}$, approximately $2.0\,\text{V}$.
$2.0\,\text{V}$.
Photon energy in electron volts is $E=hc/\lambda=1240/330=3.76\,\text{eV}$. Since $3.76\,\text{eV}<4.2\,\text{eV}$, the incident photons do not have enough energy to eject electrons from the metal.
No. The photon energy is about $3.76\,\text{eV}$, which is less than the work function $4.2\,\text{eV}$.
The maximum kinetic energy is $K_{\max}=\frac12mv^2=\frac12(9.11\times10^{-31})(6.0\times10^5)^2=1.64\times10^{-19}\,\text{J}$. Einstein's equation gives $h\nu=h\nu_0+K_{\max}$, so $\nu_0=\nu-K_{\max}/h=7.21\times10^{14}-(1.64\times10^{-19})/(6.626\times10^{-34})=4.7\times10^{14}\,\text{Hz}$.
$4.7\times10^{14}\,\text{Hz}$.
Photon energy is $E=hc/\lambda=1240/488=2.54\,\text{eV}$. Since $K_{\max}=eV_0=0.38\,\text{eV}$, the work function is $\phi=E-K_{\max}=2.54-0.38=2.16\,\text{eV}$.
$2.16\,\text{eV}$.
Use $\lambda=h/mv$. For the bullet, $p=(0.040)(1.0\times10^3)=40\,\text{kg m s}^{-1}$, so $\lambda=6.626\times10^{-34}/40=1.7\times10^{-35}\,\text{m}$. For the ball, $p=(0.060)(1.0)=0.060\,\text{kg m s}^{-1}$, so $\lambda=1.1\times10^{-32}\,\text{m}$. For the dust particle, $p=(1.0\times10^{-9})(2.2)=2.2\times10^{-9}\,\text{kg m s}^{-1}$, so $\lambda=3.0\times10^{-25}\,\text{m}$.
(a) $1.7\times10^{-35}\,\text{m}$. (b) $1.1\times10^{-32}\,\text{m}$. (c) $3.0\times10^{-25}\,\text{m}$.
A photon of electromagnetic radiation has energy $E=h\nu$ and momentum $p=E/c=h\nu/c$. Since $c=\nu\lambda$, this becomes $p=h/\lambda$. The de Broglie wavelength of the photon is therefore $\lambda_{dB}=h/p=h/(h/\lambda)=\lambda$, the wavelength of the electromagnetic radiation.
For a photon, $p=h\nu/c=h/\lambda$, so its de Broglie wavelength $h/p$ equals the electromagnetic wavelength $\lambda$.