The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h^2 \over xma_0^2 . The value of 10x is \underline\hspace2cm . (a0 is radius of Bohr's orbit) (Nearest integer) [Given : \pi = 3.14]

Structure of Atom PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q15 with Solution

View written solutionFree

Correct answer: 3155

For the hydrogen atom:

For the second orbit, $n=2$ , so $r_2 = 4a_0$

Kinetic energy of the electron is $K = \frac{1}{2}mv^2$

From Bohr's quantization, $v = \frac{nh}{2\pi m r_n}$

So,

= \frac{n^2h^2}{8\pi^2 m r_n^2}$$ Now substitute $r_n = n^2a_0$: $$K = \frac{n^2h^2}{8\pi^2 m (n^2a_0)^2} = \frac{h^2}{8\pi^2 m n^2 a_0^2}$$ For $n=2$: $$K = \frac{h^2}{8\pi^2 m (4)a_0^2} = \frac{h^2}{32\pi^2 m a_0^2}$$ 3. **Compare with the given form** Given: $$K = \frac{h^2}{xma_0^2}$$ Hence, $$x = 32\pi^2$$ 4. **Calculate $10x$** Using $\pi = 3.14$, $$x = 32(3.14)^2$$ $$x = 32(9.8596) = 315.5072$$ Therefore, $$10x = 3155.072$$ Nearest integer: $$\boxed{3155}$$ 5. **Comparison with stored answer** Derived answer = $3155$ Stored correct answer = $3155$ They match.

View written solutionFree

Correct answer: 3155

For the hydrogen atom:

For the second orbit, $n=2$ , so $r_2 = 4a_0$

Kinetic energy of the electron is $K = \frac{1}{2}mv^2$

From Bohr's quantization, $v = \frac{nh}{2\pi m r_n}$

So,

= \frac{n^2h^2}{8\pi^2 m r_n^2}$$ Now substitute $r_n = n^2a_0$: $$K = \frac{n^2h^2}{8\pi^2 m (n^2a_0)^2} = \frac{h^2}{8\pi^2 m n^2 a_0^2}$$ For $n=2$: $$K = \frac{h^2}{8\pi^2 m (4)a_0^2} = \frac{h^2}{32\pi^2 m a_0^2}$$ 3. **Compare with the given form** Given: $$K = \frac{h^2}{xma_0^2}$$ Hence, $$x = 32\pi^2$$ 4. **Calculate $10x$** Using $\pi = 3.14$, $$x = 32(3.14)^2$$ $$x = 32(9.8596) = 315.5072$$ Therefore, $$10x = 3155.072$$ Nearest integer: $$\boxed{3155}$$ 5. **Comparison with stored answer** Derived answer = $3155$ Stored correct answer = $3155$ They match.