What is the K.E and P.E of electron in the first orbit of hydrogen atom? Given e =1.6 × 10-19 C , r =0.53 × 10-10 m

Question

What is the K.E and P.E of electron in the first orbit of hydrogen atom? Given e =1.6 × 10-19 C , r =0.53 × 10-10 m (A) 13.58 eV ,-27.16 eV (B) -13.58 eV , 27.16 eV (C) 13.58 J , 27.16 eV (D) 27.16 eV , 27.16 eV

Tardigrade Admin · Accepted Answer

K.E. of electron =( KZe 2/2 r )=( Ke 2/2 r ) (∵ Z =1) K.E =(9 × 109 ×(1.6 × 10-19)2/2 × 0.53 × 10-10) Joule =(9 × 2.56 × 10-19/1.06 × 1.6 × 10-19) eV =13.58 eV P.E of electron =(- K Ze 2/ r )=-2( K , E ) =2 ×(13.58)=-27.16 eV

Q. What is the K.E and P.E of electron in the first orbit of hydrogen atom? Given e=1.6×10−19C,r=0.53×10−10 m

13.58 eV ,-27.16 eV

-13.58 eV , 27.16 eV

13.58 J , 27.16 eV

27.16 eV , 27.16 eV

K.E. of electron =2rKZe2​=2rKe2​(∵Z=1) K.E=2×0.53×10−109×109×(1.6×10−19)2​ Joule =1.06×1.6×10−199×2.56×10−19​eV=13.58eV P.E of electron =r−KZe2​=−2(K,E) =2×(13.58)=−27.16eV

Q. What is the K.E and P.E of electron in the first orbit of hydrogen atom? Given $e = 1.6 \times 1 0^{- 19} C, r = 0.53 \times 1 0^{- 10}$ $m$