The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to h 2 x m a 0 2 . The value of 10x is________. (a0 is radius of Bohr's orbit) (Nearest integer)[Given : π = 3.14]

mvr = n h 2 π ( a n g u l a r m o m e n t u m ) K E = n 2 h 2 8 π 2 m r 2 (Bohr's kinetic energy) = 4 h 2 8 π 2 m ( 4 a 0 ) 2 ( n = 2 ) K . E . = 1 3 2 π 2 h 2 m a 0 2 Comparing with h 2 x m a 0 2 x = 32π2 = 315.50 10x = 3155

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{h^{2}}{x m a_{0}^{2}} .$ The value of 10x is________. (a₀ is radius of Bohr's orbit) (Nearest integer)[Given : π = 3.14]

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Vishal Baghel | Contributor-Level 10

6 months ago

mvr = $\frac{n h}{2 π} (a n g u l a r m o m e n t u m)$

$K E = \frac{n^{2} h^{2}}{8 π^{2} m r^{2}}$ (Bohr's kinetic energy)

$= \frac{4 h^{2}}{8 π^{2} m {(4 a_{0})}^{2}} (n = 2)$

$K . E . = \frac{1}{32 π^{2}} \frac{h^{2}}{m a_{0}^{2}}$

Comparing with $\frac{h^{2}}{x m a_{0}^{2}}$

x = 32π² = 315.50

10x = 3155

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