Tiêu đề 12. ATOMS(H).pdf ✅

NEET REVISION 4 . ( ) : E x pla n a o n B = μ 0 i 2 r ( 1 ) i = e v 2 πr ( 2 ) F r o m ( 1 ) a n d ( 2 ) ⇒ B = μ 0 e v 4 πr 2 ( 3 ) F = m v 2 r = 1 4 πε 0 e 2 r 2 ⇒ v = e 2 4 πε 0 m r 1 / 2 ( 4 ) W e k n o w m v r = n h 2 π ( b y B o h r t h e o r y ) ⇒ m e 2 4 πε 0 m r 1 / 2 r = n h 2 π ⇒ m 2 e 2 4 πε 0 m r r 2 = n 2 h 2 4 π ⇒ r = n 2 h 2 ε 0 πm e 2 F r o m e q n ( 4 ) v = e 2 4 πε 0 m n 2 h 2 ε 0 πm e 2 v = e 2 2 n ε 0 h F r o m e q n ( 3 ) B = μ 0 e 4 πr 2 ⋅ e 2 2 n ε 0 h B = μ 0 e 3 × π 2 m 2 e 4 4 π 2 n ε 0 h n 2 h 4 ε 20 ⇒ B = μ 0 e 7 m 2 π 8 ε 30 h 5 5 . ( ) : E x pla n a o n 1λα = R ( Z − α ) 2 112 − 122 = 3 R4 ( Z − α ) 2 1λβ = R ( Z − α ) 2 112 − 132 = 89 R ( Z − α ) 2 ∴ λ β λ α = 2 7 3 2 ⇒ λ β = 0.2 7 A ∘ 6 . ( ) : E x pla n a o n If all the elemen t s h avi n g n > 4 a r e r e m o v e d t h e n u m b e r of ele m e n t s t h a t will b e p r e s e n t in t h e p e ‐ rio dic t a ble a r e c alc ula t e d a s ( 1 ) n = 1 , r e p r e s e n t s K s h ell a n d t h e n u m b e r of el‐ e m e n t s h avi n g K s h ell = 2 [in a c c o r d a n c e wit h 2 n 2 ] ( 2 ) n = 2 , r e p r e s e n t s L s h ell a n d t h e n u m b e r of el‐ e m e n t s h avi n g L s h ell = 8 ( 3 ) n = 3 , r e p r e s e n t s M s h ell a n d t h e n u m b e r of el‐ e m e n t s h avi n g M s h ell = 1 8 ( 4 ) n = 4 , r e p r e s e n t s N s h ell a n d t h e n u m b e r of el‐ e m e n t s h avi n g N s h ell = 3 2 S o , t h e t o t al n u m b e r of ele m e n t s h avi n g n < 5 a r e 2 + 8 + 1 8 + 3 2 = 6 0 7 . ( ) : E x pla n a o n T h e a n g ula r m o m e n t u m of dia t o mic m ole c ule is L = Iω = n h 2 π F o r S e c o n d o r bit , L = 2 h 2 π = hπ R o t a o n al e n e r g y = 12 Iω 2 = ( Iω ) 2 2 I = L 2 2 I = hπ 2 × 12I = h 2 2 Iπ 2 ( ) ( ) √ ( )( ) [ ] [ ]( )

NEET REVISION 12. () : Explana on Given, total energy given to e − in 4 th orbit, E = 15 eV Energy of nth orbit of H-atom, E ′ = − 13.6 n 2 So, energy of 4 th orbit of H-atom, E ′ = − 13.6 4 2 = − 0.85 eV ∴ Ionisa on energy of 4 th orbit = − E ′ = ( − 0.85 eV) = 0.85 eV ∴ Final energy of electron, when it comes out of H-atom, E = Total energy given to electron - Ionisa on en‐ ergy of 4th orbit, = 15 − 0.85 = 14.15 eV 13. () : Explana on ΔE = hv v = ΔE h = k 1 ( n − 1 ) 2 − 1 n 2 = k [ n 2 − n 2 − 1 + 2n ] n 2 ( n − 1 ) 2 = k2n n 2 ( n − 1 ) 2 ≈ 2k n 3 α 1 n 3 (n − 1 ≈ n) 14. () : Explana on 1 λ = R 1 n 2 1 − 1 n 2 2 ⇒ n2 = 4 ∴ Number of emission line N = n ( n − 1 ) 2 = 4 × 3 2 = 6 15. () : Explana on The wavelength of the first Balmer line is λ = 1 vˉ = 1 R 1 2 2 − 1 3 2 = 36 5Rm In general, by Doppler effect λ ′ = c − v s c − v o λ Here λ ′ = c − vcos θ c − 0 λ ∴ Δλ = λ ′ − λ = vcos θ c λ ∴ v = cΔλ λcos θ = 3 × 10 8 × 0.2 × 10 − 10 36 5R × cos 60 ∘ = 0.6 × 10 − 2 × 5 × 10 7 × 2 36 × 1 = 16.67 × 10 3 m/s ∴ N = 16.67 16. () : Explana on Bn = μ0 In 2rn Bn = μ0 e 2rnT Bn = μ0 e 2rn 2πrn vn = μ0 ev n 4πr 2 n Bn ∝ vn rn 2 But v ∝ Z n and rn ∝ n 2 Z ∴ Bn ∝ Z 3 n 5 Bn ∝ n − 5Z 3 ∴ q = 3, p = − 5 ∴ q − p = 3 − ( − 5) = 8 17. () : Explana on ΔE = hc λ = 13.6z 2 1 1 2 − 1 2 2 ⇒ λα 1 z 2 [ ] [ ] ( ) [ ] [ ] | | ( ) ( ) ( ) [ ]