If the speed of an electron of hydrogen atom in the ground state is 2.2 × 10⁶ m/s, then its speed in the third excited state will be (a) 5.5 × 10⁶ m/s (b) 5.5 × 10⁵ m/s (c) 8.8 × 10⁵ m/s (d) 6.8 × 10⁶ m/s

Answer: (b). Third excited state means n = 4; vn = v₁/n. v₄ = (2.2 × 10⁶)/(4) = 5.5 × 10⁵ m/s So the answer is (b) 5.5 × 10⁵ m/s.

If the speed of an electron of hydrogen atom in the ground state is… | MHT CET 2019 (3 May, Shift 1)

Q100MHT CET 2019 · 3 MAY · SHIFT 1Easy

Bohr Model: Orbits, Speed & Angular Momentum

If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^{6}\text{ m/s}$ , then its speed in the third excited state will be

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Answer: (b)

Third excited state means $n = 4$ ; $v_n = \frac{v_1}{n}$ .
$v_4 = \frac{2.2 \times 10^6}{4} = 5.5 \times 10^5\text{ m/s}$
So the answer is (b) $5.5 \times 10^5\text{ m/s}$ .

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Bohr Model: Orbits, Speed & Angular Momentum question from MHT CET 2019 (3 May, Shift 1)

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