Download App

2. structure of atom — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistry2. structure of atom1 Mark
MCQ
When an electron revolves in a stationary orbit then
  • A
    It absorbs energy
  • B
    It gains kinetic energy
  • C
    It emits radiation
  • ✓
    Its energy remains constant
✓

Answer

Correct option: D.
Its energy remains constant
d
When an electron revolves in a stationary orbit then its energy remain constant.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from 2. structure of atom→2. structure of atom — all question types→Chemistry STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Which reactions can be used to prepare diborane

$I.\,\,NaBH_4 + BF_3$ (in ether) $\to $

$II.\,\, NaBH_4 + I_2 \to $

$III.\,\, BF_3 + NaH \to $

→
Avogadro number is
→
The equilibrium constant K for the reaction:
2HI(g) ⇌ H2​(g) + I1​(g) at room temp is 2.85 and that at 698 K is 1.4 × 10 − 2. This implies that the forward reaction is:
→
The ionzation potential of hydrogen atom is $13.6\, eV.$ The energy required to remove an electron from the $n = 2$ state of hydrogen atom is : ............ $\mathrm{eV}$
→
Ozonolysis of  $2-$ methyl butene  $-2$  yields
→
The overall complex dissociation equilibrium constant for the complex $[Cu(NH_3)_4]^{2+}$ ion will be ( $\beta _4$ for this complex is $2.1\times 10^{13}$ ) $\beta _4 =$ association constant
→
Which of the following element have highest ionization enthalpy?
→
The dipole moment of chlorobenzene is $1.73 \,\,D$. The dipole moment of $p$-dichlorobenzene is expected to be ............. $\mathrm{D}$
→
Based on equation $E = - 2.178 \times 10^{-18} \ J \ \left( {\frac{{{Z^2}}}{{{n^2}}}} \right)$  certain conclusions are written. Which of themis not correct?
→
Product $(B)$ of the reaction is

$\begin{matrix}
   CH-C{{O}_{2}}H  \\
   \,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
   CH-C{{O}_{2}}H  \\
\end{matrix}\xrightarrow[(two\,mole)]{NaOH}(A)\xrightarrow{electrolysis}(B);$

→