C_60 emerging from a source at a speed (v) has a de Broglie wavelength of 11.0 Å . The value of v (inm s^-1 . ) is closest to [Planck's constant .h=6.626 × 10^-34 ~J ~s ]

Correct answer: (A) 0.5. λ=h/mv ⇒ v=h/(m · λ)=frac6.62 × 10^-34720 × 10^-3 × 11 × 10^-10

Q8 / 52 · KVPY 2017 (19 Nov SA) · Chemistry

$\mathrm{C}_{60}$ emerging from a source at a speed (v) has a de Broglie wavelength of $11.0 Å$. The value of $\mathrm{v}\left(\mathrm{inm} \mathrm{s}^{-1}\right.$ ) is closest to [Planck's constant $\left.h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right]$

Correct answer
A
$0.5$
B
$2.5$
C
$5.0$
D
30

Solution

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \Rightarrow \mathrm{v}=\frac{\mathrm{h}}{\mathrm{m} \cdot \lambda}=\frac{6.62 \times 10^{-34}}{720 \times 10^{-3} \times 11 \times 10^{-10}}$

Hence, the correct answer is option: 'A'.

$\mathrm{C}_{60}$ emerging from a source at a speed (v) has a de Broglie wavelength of $11.0 Å$. The value of $\mathrm{v}\left(\mathrm{inm} \mathrm{s}^{-1}\right.$ ) is closest to [Planck's constant $\left.h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right]$

Solution

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