The time taken for an electron to complete one revolution in Bohr

Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

The boiling points of HF, HCl, HBr and HI follow the order

  • HF> HCl > HBr > HI

  • HF> HI> HBr > HCl

  • HI> HBr > HCl> H

  • HCl > HF> HBr > HI


2.

The energy required to break one mole of hydrogen-hydrogen bonds in H2 is 436 kJ. What is the longest wavelength of light required to break a single hydrogen-hydrogen bond?

  • 68.5nm

  • 137nm

  • 274nm

  • 584nm


3.

The correct order of O-O bond lenght in O2, H2O2, and O3

  • O2> O3> H2O2

  • H2O2 > O3 > O2

  • O3> O2> H2O2

  • O3 > H2O2> O2


4.

The number of σ and π bonds between two carbon atoms in calcium carbide are

  • one σ, one π

  • one σ, two π

  • two σ, one π

  • one σ, 112π


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5.

An element E loses one α and two β particles in three successive stages. The resulting elemenl wiil be

  • an isobar of E

  • an isotone of E

  • an isotope of E

  • E itself


6.

An element X belongs to fourth period and fifteenth group of the periodic table. Which of the following statements is true?

  • It has completely filled s-orbital and a partially filled d-orbital.

  • It has completely filled s-and p -orbitals and a partially filled d-orbital.

  • It has completely filled s- and p-orbitals and a half filled d-orbital.

  • It has a half filled p -orbital and completely filled s-and d-orbitals.


7.

Which of the following plots represent an exothermic reaction?


8.

The condition for a reaction to occur spontaneously is

  • H must be negative

  • S must be negative

  • H-TS must be negative

  • H+TS must be negative


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9.

The molar solubility (in mol L-1) of a sparingly soluble salt MX4 is 'S'. The corresponding solubility product is Ksp in terms of 'Ksp' is given by the relation

  • S=Ksp1281/4

  • S=Ksp2561/5

  • S=(256 Ksp)1/5

  • S=(128 Ksp)1/4


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10.

The time taken for an electron to complete one revolution in Bohr orbit of hydrogen atom is

  • 4m2πr2n2h2

  • n2h24mr2

  • 4π2mr2nh

  • nh4π2mr2


C.

4π2mr2nh

Let the distance travelled in Ttime = 2πr                                    (Circumference of orbit)Velocity(v) =2πrT........ (i) also v, = n·h2πmr.........(ii)therefore,            1T(Time period)= frequency ν= v×12πrFrom Eq. (ii), ν=1T=nh2πr×2πmr= nh4π2r2mT (time taken in one revolution)  =4π2r2mnh


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