Question

a hydrogen atom undergoes an electronic transition from the n = 4 to the n = 2 state. in the process, the h atom emits a photon, which then strikes a cesium metal surface and ejects an electron. it takes 3.23×10^(-19) j to remove an electron from cs metal. calculate (a) the energy of the n = 4 state of the h atom. u_4 = i×10 i j (b) the wavelength of the emitted photon. λ = i nm (c) the energy of the ejected electron. u_electron = i×10 i j (d) the wavelength of the ejected electron. λ = i nm

Explanation:

Step1: Calculate energy of n = 4 state of H - atom

The energy of the n - th state of a hydrogen atom is given by $E_n=-\frac{13.6}{n^{2}}\text{ eV}$. First convert from eV to J. $1\text{ eV}=1.6\times 10^{-19}\text{ J}$. For $n = 4$, $E_4=-\frac{13.6}{4^{2}}\text{ eV}=-\frac{13.6}{16}\text{ eV}=- 0.85\text{ eV}$. In joules, $E_4=-0.85\times1.6\times 10^{-19}\text{ J}=-1.36\times 10^{-19}\text{ J}$.

Step2: Calculate wavelength of the emitted photon

The energy of the photon emitted during the transition from $n = 4$ to $n = 2$ is $\Delta E=E_4 - E_2$. $E_2=-\frac{13.6}{2^{2}}\text{ eV}=- 3.4\text{ eV}$. $\Delta E=(-0.85)-(-3.4)\text{ eV}=2.55\text{ eV}$. In joules, $\Delta E = 2.55\times1.6\times 10^{-19}\text{ J}=4.08\times 10^{-19}\text{ J}$. Using the formula $E = h
u=\frac{hc}{\lambda}$, where $h = 6.63\times 10^{-34}\text{ J}\cdot\text{s}$ and $c = 3\times 10^{8}\text{ m/s}$. Then $\lambda=\frac{hc}{\Delta E}=\frac{6.63\times 10^{-34}\times3\times 10^{8}}{4.08\times 10^{-19}}\text{ m}\approx487\times 10^{-9}\text{ m}=487\text{ nm}$.

Step3: Calculate energy of the ejected electron

The energy of the ejected electron is given by the photoelectric effect $K_{e}=E_{photon}-\phi$, where $\phi = 3.23\times 10^{-19}\text{ J}$ and $E_{photon}=4.08\times 10^{-19}\text{ J}$. So $K_{e}=4.08\times 10^{-19}-3.23\times 10^{-19}\text{ J}=0.85\times 10^{-19}\text{ J}$.

Step4: Calculate wavelength of the ejected electron

The de - Broglie wavelength of the electron is given by $\lambda=\frac{h}{p}$, and $p=\sqrt{2mK}$, where $m = 9.1\times 10^{-31}\text{ kg}$ and $K = 0.85\times 10^{-19}\text{ J}$. First, $p=\sqrt{2\times9.1\times 10^{-31}\times0.85\times 10^{-19}}\text{ kg}\cdot\text{m/s}\approx1.24\times 10^{-24}\text{ kg}\cdot\text{m/s}$. Then $\lambda=\frac{h}{p}=\frac{6.63\times 10^{-34}}{1.24\times 10^{-24}}\text{ m}\approx5.35\times 10^{-10}\text{ m}=0.535\text{ nm}$.

Answer:

(a) $-1.36\times 10^{-19}$
(b) $487$
(c) $0.85\times 10^{-19}$
(d) $0.535$