please please solve the above question.whose answer is 2.19×10^6 m/s
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circumstances of 1st orbit = 2π × radius of 1st orbit = 3.332 × 10^-10 m
2 × 22/7 × radius of 1st orbit = 3.332 × 10^-10 m
radius of 1st orbit = (3.332/6.28 ) × 10^-10
= (0.529) × 10^-10
now, use ,
mvr = nh/2π { according to Bohr theory
v = h/2πmr { n = 1 ( 1st orbit
put value of h= 6.626 × 10^-34 , m = 9.1 × 10^-31 kg ,r = 0.529 × 10^-10
v = 2.188× 10^6 m/s
≈ 2.19 × 10^6 m/s (approximately)
2 × 22/7 × radius of 1st orbit = 3.332 × 10^-10 m
radius of 1st orbit = (3.332/6.28 ) × 10^-10
= (0.529) × 10^-10
now, use ,
mvr = nh/2π { according to Bohr theory
v = h/2πmr { n = 1 ( 1st orbit
put value of h= 6.626 × 10^-34 , m = 9.1 × 10^-31 kg ,r = 0.529 × 10^-10
v = 2.188× 10^6 m/s
≈ 2.19 × 10^6 m/s (approximately)
anil43:
its 2nd orbit of hydrogen atom
okay okay i understand thx for your kind help
no this velocity of electron in 1st orbit
welcome
abhi you are so good in studies is there any way by which i can ask my questions to you?????
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