Sebuah elektron energi totalnya n kali energi diamnya

Sebuah elektron energi totalnya n kali energi diamnya. Jika massa diam elektron adalah mo, konstanta Planck h adalah dan kelajuan cahaya di ruang hampa adalah c, maka panjang gelombang de Broglie elektron tersebut adalah

A
$$\frac{h\sqrt{n^{2}-1}}{m_{0}c}$$
B
$$\frac{h\sqrt{n^{2}+1}}{m_{0}c}$$
C
$$\frac{h}{n.m_{0}c}$$
D
$$\frac{h}{\sqrt{n^{2}-1}.m_{o}c}$$
E
$$\frac{h}{\sqrt{n^{2}+1}.m_{o}c}$$

$$\frac{h}{\sqrt{n^{2}-1}.m_{o}c}$$

Panjang gelombang de Broglie:
$$\lambda =\frac{h}{mv}$$

dengan massa relativistik :
$$\begin{aligned}&m=\frac{m_{o}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\gamma m_{o}\\&\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\end{aligned}$$

energi total dari benda yang bergerak: $$E=mc^{2}$$
energi diam benda:$$E_{o}=m_{o}c^{2}$$

maka:
$$\begin{aligned}&E=mc^{2}=\left (\gamma m_{o} \right )c^{2}=\gamma\left ( m_{o} c^{2} \right )=\gamma E_{o}\end{aligned}$$

energi total elektron n kali energi diamnya, maka:
$$\begin{aligned}&E=nE_{o}\\&\gamma E_{o}=nE_{o}\\&\gamma=n\\\\&n = \gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\&n^{2} = \frac{1}{1-\frac{v^{2}}{c^{2}}}\\&n^{2} = \frac{c^{2}}{c^{2}-v^{2}}\\&\frac{c^{2}}{n^{2}} = c^{2}-v^{2}\\&v^{2} = c^{2} – \frac{c^{2}}{n^{2}}\\&v^{2} = \frac{\left ( n^{2}-1 \right )c^{2}}{n^{2}}\\&v = \sqrt{\frac{\left ( n^{2}-1 \right )c^{2}}{n^{2}}}\\&v = \frac{\sqrt{n^{2}-1}}{n}c^{2}\end{aligned}$$

sehingga:
$$\begin{aligned}&\lambda = \frac{h}{mv} =\frac{h}{\left ( \gamma m_{o} \right )\left ( \frac{\sqrt{n^{2}-1}}{n}c\right)}\\&=\frac{h}{n.m_{o}.\frac{\sqrt{n^{2}-1}}{n}c}\\&=\frac{h}{\sqrt{n^{2}-1}.m_{o}c}\\\end{aligned}$$

Source: SBMPTN 2014 | SAINTEK | FISIKA | Kode soal 512 | 19
Topik: Teori Relativitas