Title DPP-7 SOLUTION.pdf ✅

CLASS : XIth SUBJECT : CHEMISTRY DATE : DPP No. : 7 1 (a) λ for visible light is in the range of 400 to 780 nm. E = hc λ . This, it is in the range of electron volt (eV). 3 (a) To cross over threshold energy level. 4 (d) ∆E = hv = hc λ λ = hc ∆E = 6.62 × 10―34 × 3 × 108 4.4 × 10―14 = 4.52 × 10―12m 5 (c) r2Be 3+ = r1H 4 × 2 2 ( ∵ r2H = r1 H × 2 2 and rnBe 3+ = rnH n ) 6 (b) An experimental fact. 7 (d) The transition is almost instantaneous process 8 (b) The values of m are ―l to +l through zero. 9 (b) A fact. 10 (c) X-rays are light waves or a form of light energy. 11 (c) ∆x ∙ ∆v ≥ h 4πm 12 (d) Topic :- STRUCTURE OF ATOM Solutions
v = 1 λ = R′Z 2 [ 1 n 2 1 ― 1 n 2 2 ] For shortest wavelength (maximum energy) in Lyman series of hydrogen Z = 1,n1 = 1,n2 ⟶∞ and λ = x 1 x = R′ For longest wavelength (minimum energy) in Balmer series of He +,Z = 2 and n1 = 2,n2 = 3 1 λ = R′2 2 [ 1 2 2 ― 1 3 2] 1 λ = 4 x [ 1 4 ― 1 9 ] 1 λ = 4 x 5 36 λ = 9x 5 13 (d) Rydberg is an unit of energy. 14 (a) Neutrons are neutral particles. 15 (d) + 1 2 and ― 1 2 spinning produces angular momentum equal to Z ― component of angular momentum which is given as ms (h/2π) 16 (c) Since, hv = work function + (1/2)mu 2 . 17 (d) λ = h p v = c λ v = 3 × 108 × 1.1 × 10―23 6.6 × 10―34 = 5.0 × 1018Hz 18 (b) E = hc λ = hv 19 (b) Step 1 Calculate energy given to I2 molecule by hc λ Step 2 Calculate energy used to break I2 molecule. The difference in above two energies will be the KE of two I atoms 20 (a) It is a fact.
ANSWER-KEY Q. 1 2 3 4 5 6 7 8 9 10 A. A C A D C B D B B C Q. 11 12 13 14 15 16 17 18 19 20 A. C D D A D C D B B A