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Grade: 11

                        

The kinetic energy of an electron in H like atom is 6.04eV. The area of the third Bohr orbit to which this electron belong is1. 1.78×10^-16 cm^22. 17.80×10^-16 cm^23. 35.6×10^-16 cm^24. 3.56×10^-16 cm^2

2 years ago

Answers : (1)

Arun
24742 Points
							
Dear Student
 
K.E. = 6.04 ev = 9.7 * 10^-19 J
K.E. of electron = Kze²/2r
K = constant = 9* 10^9 Nm²/C²
z = 1( for hydrogen like atom)
e = 1.6 * 10^-19 C
Hence put all this in the above formula
You will get
r = 1.19 * 10^-10 metres
Hence area A = (3.14159) * r²
A= 17.8 * 10^-20 m²
Or 
A = 17.8 * 10^-16 cm² 
Hence option 2 is correct.
 
Regards
Arun (AskIITians Forum Expert)
2 years ago
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