Orbital angular momentum associated with 2p-electron is:
A. `sqrt(2)h)/(pi)`
B. `0`
C. `sqrt(6)xx(h)/(pi)`
D. `(h)/sqrt(2pi)`
Correct Answer - B `(E_(e))/(E_(p))=((1)/(2)mv^(2))/(hupsilon)=(1)/(2)vxx(mv)/(h).(h)/(hupsilon)` But `(h)/(mv)=(h)/(hupsilon)` `:.(E_(e))/(E_(ph))=(1)/(2)v[(mv)/(h).(h)/(hupsilon).(c)/(c)]=(v)/(2c)`
2 Answers 1 viewsCorrect Answer - A `L=(nh)/(2pi)` it does not denpend on Z
2 Answers 1 viewsCorrect Answer - A `lambda (h)/(p)=(h)/(mv)` or `" " lambda prop (1)/(v)` `:. " " lambda_(n)prop n rArr J_(n)=n(h)/(2pi) i.e., J_(n) prop n` Hence, `" " lambda_(n)propJ_(n)`
2 Answers 1 viewsCorrect Answer - B `L=n((h)/(2pi))i.e.,Lpropn.` In ground state n=1 so L= minimum.
2 Answers 4 viewsCorrect Answer - A Here, `E=-13.6eV=-13.6xx1.6xx10^(-19)=-2.2xx10^(-18)J` `E=(-e^2)/(8piepsilon_0r)` `therefore` As orbital radius, `r=(-e^2)/(8piepsilon_0E)=(9xx10^(9)xx(1.6xx10^(-19))^2)/(2xx(2.2xx10^(-18)))=5.3xx10^(-11)m`.
2 Answers 1 viewsCorrect Answer - `v_(e)=1836 v_(p)` Debroglie wavelength associated with particle of mass `(m)` moving with velocity `(v)` is `lambda=h/(mv) rArr lambda_(P)=h/(m_(P)v_(P))` and `lambda_(e)=h/(m_(e)v_(e))` given `lambda_(p)=lambda_(e) rArr (h)/(m_(p)v_(p))=h/(m_(e)v_(e)) rArr m_(p)v_(p)=m_(e)v_(e)` `v_(e)/v_(p)=m_(p)/m_(e)=1836 rArr...
2 Answers 1 viewsCorrect Answer - A Permissible value of orbit angular momentum of `e^(-)=n((h)/(2pi))` n = 1, 2, 3………… for n = 3, orbit angular momentum = `1.5 ((h)/(pi))`
2 Answers 1 viewsCorrect Answer - D Cation at `3^(@) gt 2^(@) gt 1^(@)`
2 Answers 1 viewsCorrect answer is (A) L = nh/2π
2 Answers 1 viewsCorrect Answer - `6.613 Å`
2 Answers 3 views