question 15 of 20\nyour answer is partially correct.\na hydrogen atom undergoes an electronic transition…

question 15 of 20\nyour answer is partially correct.\na hydrogen atom undergoes an electronic transition from the n = 4 to the n = 2 state. in the process, the h then strikes a cesium metal surface and ejects an electron. it takes 3.23×10^(-19) j to remove an electron from\n(a) the energy of the n = 4 state of the h atom.\nu4 = -1.36×10^(-19) j\n(b) the wavelength of the emitted photon.\nλ = 487 nm\n(c) the energy of the ejected electron.\nuelectron = 0.85×10^(-19) j\n(d) the wavelength of the ejected electron.\nλ = 0.535 nm

question 15 of 20\nyour answer is partially correct.\na hydrogen atom undergoes an electronic transition from the n = 4 to the n = 2 state. in the process, the h then strikes a cesium metal surface and ejects an electron. it takes 3.23×10^(-19) j to remove an electron from\n(a) the energy of the n = 4 state of the h atom.\nu4 = -1.36×10^(-19) j\n(b) the wavelength of the emitted photon.\nλ = 487 nm\n(c) the energy of the ejected electron.\nuelectron = 0.85×10^(-19) j\n(d) the wavelength of the ejected electron.\nλ = 0.535 nm

Answer

Explanation:

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

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

Step2: Calculate wavelength of emitted photon

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

Step3: Calculate energy of ejected electron

The energy of the ejected electron is the energy of the photon minus the work - function of cesium. The energy of the photon is $4.08\times 10^{-19}\text{ J}$ and the work - function $\Phi=3.23\times 10^{-19}\text{ J}$. So $U_{electron}=4.08\times 10^{-19}-3.23\times 10^{-19}\text{ J}=0.85\times 10^{-19}\text{ J}$.

Step4: Calculate wavelength of ejected electron

The energy of the ejected electron $U_{electron}=0.85\times 10^{-19}\text{ J}$. First, find the momentum $p$ of the electron using $U_{electron}=\frac{p^{2}}{2m}$ (where $m = 9.11\times 10^{-31}\text{ kg}$). So $p=\sqrt{2mU_{electron}}=\sqrt{2\times9.11\times 10^{-31}\times0.85\times 10^{-19}}\text{ kg}\cdot\text{m/s}\approx1.24\times 10^{-24}\text{ kg}\cdot\text{m/s}$. Then, using the de - Broglie wavelength formula $\lambda=\frac{h}{p}$, we get $\lambda=\frac{6.63\times 10^{-34}}{1.24\times 10^{-24}}\text{ m}\approx0.535\times 10^{-9}\text{ m}=0.535\text{ nm}$.

Answer:

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