CUET · PHYSICS · PYQ PAPER 2025
In a Young's double-slit experiment, the slits are separated by 0.56 mm and the screen is placed 2.8 m away from the slits. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
- A 600 nm
- B 400 nm
- C 800 nm
- D 700 nm
Answer & Solution
Correct Answer
(A) 600 nm
Step-by-step Solution
Detailed explanation
\(\lambda = \frac{y_n d}{n L}\) \(\lambda = \frac{(1.2 \times 10^{-2} \text{ m})(0.56 \times 10^{-3} \text{ m})}{(4)(2.8 \text{ m})}\) \(\lambda = 6.0 \times 10^{-7} \text{ m}\) \(\lambda = 600 \text{ nm}\)
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The transition elements which give the greatest number of oxidation states occur in or near the middle of the series. Manganese, for example, exhibits all the oxidation states from +2 to +7 . The lesser number of oxidation states at extreme ends stems from either too few electrons to lose or share (Sc, Ti) on too many d-electrons (hence fewer orbitals available in which to share electrons with others) for higher valence \(( Cu , Zn )\). Thus early in the series scandium (II) is virtually unknown and titanium (IV) is more stable than Ti (III) or Ti (II). At the other end, the only oxidation state of zinc is \(+ 2\) (no d electrons are involved). The maximum oxidation states of reasonable stability correspond in value to the sum of the s - and d-electrons upto manganese ( \(Ti ^{I V} O _2, V^V O _2^{+}, Cr ^{V I} O _4^{2-}\), \(Mn ^{\text {VII }} O _4\) ) followed by a rather abrupt decrease in stability of higher oxidation states, so that the typical species to follow are \(Fe ^{I I, I I I}, Co ^{I I, I I I}, Ni ^{I I}, Cu ^{I, I I}, Zn ^{I I}\).
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