What atomic number of an element "X" would have to become so that the 4th orbit around X would fit inside the I Bohr orbit of H atom ?

Correct Answer - A::C Angular momentum `=(nh)/(2pi)=(3h)/(2pi)implies n=3` Also `r=n^(2)/2 a_(0) implies (9a_(0))/2=3^(2)/Z a_(0) implies Z=2` For de-excitation `1/lambda=Rz^(2) [1/n_(1)^(2)-1/n_(2)^(2)]=4R[1/n_(1)^(2)-1/n_(2)^(2)]` For `n=3` to `n=1` : `1/lambda=4R[1/1-1/9] implies lambda=9/(32R)` For `n=3` to...

Correct Answer - B `L=n((h)/(2pi))i.e.,Lpropn.` In ground state n=1 so L= minimum.

Correct Answer - A,B,D `L_1/L_2=(mv_1r_1)/(mv_2r_2)=(Z_1/n_1xxn_1^2/Z_1)/(Z_2/n_2xxn_2^2/Z_2)=n_1/n_2implies P_1/P_2(mv_1)/(mv_2)=(Z_1/n_1)/(Z_2/n_2)=(Z_1n_2)/(Z_2n_1)` `(f_1)/(f_2) =(v_1/(2pir_1))/(v_2/(2pir_2))=(Z_1/n_1xxZ_1/n_1^2)/(Z_2/n_2xxZ_2/n_2^2)=(Z_1/Z_2)^2.(n_2/n_1)^3implies (K.E._1)/(K.E._2)=(1/2mv_1^2)/(1/2mv_2^2)=(Z_1/Z_2)^2xx(n_2/n_1)^2=((Z_1n_2)/(Z_2n_1))^2` `(K.E_1)/(K.E_2)=(1/2(KZ_1e^2)/r_1)/(1/2(KZ_2e^2)/r_2)=(Z_1/n_1^2.Z_1)/(Z_2.Z_2/n_2^2)=(Z_1/Z_2.n_2/n_1)^2`

Correct Answer - A In the ground state of hydrogen atom, suppose, `a_0=` bohr radius `v_0=` velocity of electron in first orbit `therefore` time taken by electron to complete one revolution,...

Correct Answer - C As`r_n=n^2a_0`, Here, `a_0=0.529Å` and `n=3` `therefore r_(n=3)=(3)^2a_0=9xx0.529Å=4.761Å`

Correct Answer - B `because 2pir_n=nlambda` `therefore n=2,2pir_2=2lambda=2xx`de broglie wavelength.

Correct Answer - A The element which present just below the given element will have outermost electromic configuration as `4s^(2) 4p^(3),` so its full electronic configuration is `1s^(2),2s^(2),2p^(6),3s^(2)3p^(6),4s^(2), 3d^(10), 4p^(3)` and...

Correct Answer - D `r_(n)=0.529 n^(2)/Z ` For hydrogen, `n=1` and `Z=1, :. r_(H)=0.529` For `Be^(3+), n=2` and `Z=4, :. r_(Be^(3+))=(0.529xx2^(2))/(4)=0.529` there, (D) is correct option.

Option : C. 16 

Correct Answer - (b) `z = 48`