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Current Question (ID: 7498)

Question:

The wavelength of the radiation emitted when in a H atom, the electron falls from infinity to stationary state (n=1), is:

Options:

1. 15 $nm$
2. 192 $nm$
3. 406 $nm$
4. 91 $nm$

Solution:

### **Hint:** Apply the Rydberg formula for hydrogen atom transitions **Step 1:** Use the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \] where $R_H$ is the Rydberg constant for hydrogen ($1.096773 \times 10^7 m^{-1}$). **Step 2:** For transition from $n_2 = \infty$ to $n_1 = 1$: \[ \frac{1}{\lambda} = R_H \left[ 1 - 0 \right] = R_H \] \[ \lambda = \frac{1}{R_H} \] **Step 3:** Calculate the wavelength: \[ \lambda = \frac{1}{1.096773 \times 10^7} m = 9.12 \times 10^{-8} m = 91.2 nm \] **Step 4:** The closest option is 91 nm. Thus, the correct answer is 91 nm.

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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
}