হাইড্রোজেন পরমাণুর মধ্যে ই লেকট্রন প্রোটনের চারপাশে `2.18 xx 10^(6) গতির বৃত্তে

(1)/(1*6xx10^(-18))

Calculate thevelocity of the electron in the third orbit of hydrogen atom Hint : v= ( 2.18 xx 10^(6)z)/( n) = m//s

The electron in first orbit of hydrogen with velocity 2.18 xx 10^(6) m//s . If the radius of first orbit is 0.53 Å, then orbital current is

The frequency of radiation emiited when the electron falls n =4 to n=1 in a hydrogen atom will be ( given ionization energy of H= 2.18 xx 10 ^(-18)J "atom "^(-1) and h= 6.625 xx 10 ^(-34)Js)

The wave number of two waves are 2.5 xx 10^(6) / metre and 2 xx 10^(6) /metre. What is the difference between their wavelengths ?