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Current Question (ID: 7484)

Question:

$\text{The number of photons of light with a wavelength of 4000 pm that provide 1J of energy would be:}$

Options:

1. $2.01 \times 10^{16}$
2. $2.01 \times 10^{19}$
3. $4.14 \times 10^{23}$
4. $2.14 \times 10^{21}$

Solution:

$\text{HINT: Use the expression to find energy of photons i.e. Energy (E}_n\text{) of 'n' photons = nh}\nu$ $\text{STEP 1: Write down the given data}$ $\text{Given,}$ $\lambda \text{ (wavelength of light)} = 4000 \text{ pm} = 4000 \times 10^{-12} \text{ m}$ $c \text{ (velocity of light in vacuum)} = 3.0 \times 10^8 \text{ ms}^{-1}$ $h \text{ (Planck's constant)} = 6.626 \times 10^{-34} \text{ Js}$ $\text{STEP 2: Find out the number of photon using energy}$ $\text{We know that}$ $\text{Energy (E) of a photon} = h\nu$ $\text{Energy (E}_n\text{) of 'n' photons} = nh\nu$ $\Rightarrow n = \frac{E_n\lambda}{hc}$ $\text{Substituting the values in the expression of n;}$ $n = \frac{1 \text{J} \times 4000 \times 10^{-12} \text{m}}{6.626 \times 10^{-34} \text{Js} \times 3.0 \times 10^8 \text{ms}^{-1}} = 2.01 \times 10^{16}$ $\text{Hence, the number of photons with a wavelength of 4000 pm and energy of 1J are } 2.01 \times 10^{16}$

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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
}