find the circumference of the fifth bohr orbit of an electron whose de-brogile wavelength corresponds to translation kinetic energy of h at 27 degree celsius

Translational kinetic energy = 32kTk is called Boltzmann constant, whose value is = 1.38064852 × 10-23 T= 27 0C = 300 KSo Translational kinetic energy = 32kT = 32 × 1.3806× 10-23  × 300                               = 1242.54 × 10-23This translational kinetic energy = Wavelength λCircumference 2πr = where n is the number of orbitCircumference 2πr = 5 × 1242.54 × 10-23 = 6212.7 × 10-23 m = 6.212 × 10-18 cm

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