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Standard XII

Chemistry

De Broglie's Hypothesis

If H 2, He,...

Question

If $H_{2}$ , He, $O_{2}$ $N_{2}$ travel with same velocity then the longest de-Broglie wavelength should be associated with:

H_{2}

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H e

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O_{2}

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N_{2}

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Solution

The correct option is A $H_{2}$
De-broglie wavelength of an electron is given by,
$λ = \frac{h}{m v}$
Here, $v$ and $h$ is constant
Therefore, $λ$ $α$ $\frac{1}{m}$
Hence mass of hydrogen is least and its wavelength is longest.

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Similar questions

Q. If

H_{2}

He,

O_{2}

N_{2}

travel with the same velocity then the longest de-broglie wavelength should be associated with

Q. If

H_{2}, H e, O_{2}

and

N_{2}

tarvel with same velocity then the longest de-Brogile wavelength should be associated with:

Q. Gases

H_{2} H e O_{2} N_{2}

Critical temperature(K) 33.2 5.3 154.3 126

From the above data what whould be the order of liquefaction of these gases?
Start writing the order from the gas liquefying first.

Gases posses charcteristic critical temperature which depends upo the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.

$\begin{matrix} G a s e s & H_{2} & H e & O_{2} & N_{2} C r i t i c a l t e m p e r a t u r e i n K e l v i n & 33.2 & 5.3 & 154.3 & 126 \end{matrix}$

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first

(a) $H_{2}, H e, O_{2}, N_{2}$ (b) $H e, O_{2}, H_{2}, N_{2}$

Q. At

S T P

the order of mean square velocity of molecules of

H_{2}, N_{2}, O_{2}

and

H B r

is-

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If H2, He, O2 N2 travel with same velocity then the longest de-Broglie wavelength should be associated with:

The correct option is A H2De-broglie wavelength of an electron is given by,λ=hmvHere, v and h is constant Therefore, λ α 1mHence mass of hydrogen is least and its wavelength is longest.

If $H_{2}$ , He, $O_{2}$ $N_{2}$ travel with same velocity then the longest de-Broglie wavelength should be associated with:

The correct option is A $H_{2}$
De-broglie wavelength of an electron is given by,
$λ = \frac{h}{m v}$
Here, $v$ and $h$ is constant
Therefore, $λ$ $α$ $\frac{1}{m}$
Hence mass of hydrogen is least and its wavelength is longest.