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Current Question (ID: 7506)

Question:

The energy of an electron in an $\text{H}$-atom is given by $E_n = (-2.18 \times 10^{-18})/\text{n}^2 \text{ J}$. The shortest wavelength of light that can be used to remove an electron completely from $\text{n} = 2$ orbit will be:

Options:

1. $3647 \mathring{\text{A}}$
2. $5132 \mathring{\text{A}}$
3. $3017 \mathring{\text{A}}$
4. None of these

Solution:

Hint: Use the energy expression of a hydrogen atom to calculate wavelength. Step 1: From the given expression calculate $\Delta E$. Given, $E_n = (-2.18 \times 10^{-18})/\text{n}^2 \text{ J}$ The energy required for ionization from $\text{n}=2$ is the energy difference between the electron being completely removed (i.e., at $\text{n}=\infty$) and its initial state at $\text{n}=2$. This means the electron transitions from $\text{n}=2$ to $\text{n}=\infty$. $\Delta E = E_{\infty} - E_2$ Since $E_{\infty} = (-2.18 \times 10^{-18})/\infty^2 = 0 \text{ J}$, and $E_2 = (-2.18 \times 10^{-18})/2^2 \text{ J}$, we have: $\Delta E = 0 - \left( \frac{-2.18 \times 10^{-18}}{2^2} \right) \text{ J}$ $\Delta E = - \left( \frac{-2.18 \times 10^{-18}}{4} \right) \text{ J}$ $\Delta E = 0.545 \times 10^{-18} \text{ J}$ $\Delta E = 5.45 \times 10^{-19} \text{ J}$ Step 2: Calculate $\lambda$ from the energy and wavelength relationship. We know that the energy of a photon (which causes the transition) is given by: $\Delta E = \frac{\text{hc}}{\lambda}$ Here, $\lambda$ is the shortest wavelength causing the transition because the energy $\Delta E$ is the minimum energy required to remove the electron (ionization energy from $\text{n}=2$). A shorter wavelength would correspond to higher energy, which would also remove the electron. However, the question asks for the wavelength *that can be used to remove an electron completely*, which implies the minimum energy needed. Rearranging the formula to solve for $\lambda$: $\lambda = \frac{\text{hc}}{\Delta E}$ Where: $\text{h}$ (Planck's constant) $= 6.626 \times 10^{-34} \text{ J s}$ $\text{c}$ (speed of light) $= 3 \times 10^8 \text{ m/s}$ Substitute the values: $\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}{5.45 \times 10^{-19} \text{ J}}$ $\lambda = \frac{1.9878 \times 10^{-25}}{5.45 \times 10^{-19}} \text{ m}$ $\lambda \approx 3.647 \times 10^{-7} \text{ m}$ To convert meters to Angstroms ($\mathring{\text{A}}$), we use the conversion factor $1 \mathring{\text{A}} = 10^{-10} \text{ m}$: $\lambda = 3.647 \times 10^{-7} \text{ m} \times \frac{1 \mathring{\text{A}}}{10^{-10} \text{ m}}$ $\lambda = 3.647 \times 10^{3} \mathring{\text{A}}$ $\lambda = 3647 \mathring{\text{A}}$ The final answer is $\boxed{1}$

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{
  "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
}