Import Question JSON

Current Question (ID: 20880)

Question:

$\text{The energy of an electron is given by } E = -2.178 \times 10^{-18} \text{ J} \left( \frac{z^2}{n^2} \right)$ $\text{Wavelength of light required to excite an electron in a hydrogen atom from level } n = 1 \text{ to } n = 2 \text{ will be:}$ $(h = 6.62 \times 10^{-34} \text{ Js and } c = 3.0 \times 10^8 \text{ ms}^{-1})$

Options:

1. $6.500 \times 10^{-7} \text{ m}$
2. $8.500 \times 10^{-7} \text{ m}$
3. $1.214 \times 10^{-7} \text{ m}$
4. $2.816 \times 10^{-7} \text{ m}$

Solution:

$\text{Hint: Use the formula, that is, } 2.18 \times 10^{-18} \text{ J} \left( \frac{1}{n_1^2} - \frac{1}{n_f^2} \right) Z^2$ $\text{Step 1: Calculate the value of energy for } n = 1, \text{ and } n = 2 \text{ shell.}$ $E_1 = -2.178 \times 10^{-18}$ $E_2 = -2.178 \times 10^{-18} \times \left( \frac{1}{4} \right)$ $\Delta E = |E_2 - E_1| = 2.178 \times 10^{-18} \times \frac{3}{4}$ $\text{Step 2: Calculate the value of wavelength as follows:}$ $\Delta E = \frac{hc}{\lambda}$ $2.178 \times 10^{-18} \times \frac{3}{4} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{\lambda}$ $\lambda = \frac{19.8 \times 10^{-26} \times 4}{2.178 \times 3 \times 10^{-18}}$ $= 12.12 \times 10^{-8} \text{ m} = 1.21 \times 10^{-7} \text{ m}$ $\text{Thus, option third is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}