Question

The value of \[ \frac{{}^{100}C_{50}}{51} + \frac{{}^{100}C_{51}}{52} + \cdots + \frac{{}^{100}C_{100}}{101} \] is:

• A. \( \dfrac{2^{100}}{100} \)
• B. \( \dfrac{2^{101}}{101} \)
• C. \( \dfrac{2^{100}}{101} \)
• D. \( \dfrac{2^{101}}{100} \)

Hint:

Whenever a binomial term is divided by \(r+1\), try converting it into a higher combination using \(\displaystyle \frac{{}^nC_r}{r+1} = \frac{{}^{n+1}C_{r+1}}{n+1}\).

Answer 1

The Correct Option is C

To solve this problem, we calculate the energies of the individual levels first and then find the energy released during the transition.

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Step 1: Write the energy level formula for hydrogen-like ions

E_n = −13.6 × Z² / n²   (eV)

For the helium ion He⁺, the atomic number is:

Z = 2

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Step 2: Calculate the energy of the initial level (n = 2)

E₂ = −13.6 × 2² / 2²

E₂ = −13.6 eV

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Step 3: Calculate the energy of the final level (n = 1)

E₁ = −13.6 × 2² / 1²

E₁ = −54.4 eV

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Step 4: Calculate the energy released during the transition (2 → 1)

Energy released is the difference between final and initial energies:

ΔE = E₁ − E₂

ΔE = (−54.4) − (−13.6)

ΔE = −40.8 eV

The negative sign indicates energy emission. Hence, the magnitude of energy released is:

40.8 eV

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Final Answer:

Energy released in the transition He⁺ (2 → 1) = 40.8 eV