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Standard XII

Physics

Matter Waves

Figure shows ...

Question

Figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for the tube A in which the potential difference between the target and the filament is $V_{A}$ and the atomic number of the target material is $Z_{A}$ . These quantities are $V_{B}$ and $Z_{B}$ for the other tube. Then

V_{A} > V_{B}, Z_{A} > Z_{B}

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V_{A} > V_{B}, Z_{A} < Z_{B}

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V_{A} < V_{B}, Z_{A} > Z_{B}

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V_{A} < V_{B}, Z_{A} < Z_{B}

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Solution

The correct option is B $V_{A} > V_{B}, Z_{A} < Z_{B}$
$λ_{m i n} o f A < λ_{m i n} o f B$
so $V_{A} > V_{B}$
also $λ_{K_{α, A}} > λ_{K_{α, B}}$
so $Z_{B} > Z_{A}$

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Similar questions

Q. The figure shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for the tube A in which the potential difference between the target and the filament is V_A and the atomic number of the target material is Z_A. These quantities are V_B and Z_B for the other tube. Then,
(a) V_A > V_B, Z_A > Z_B
(b) V_A > V_B, Z_A < Z_B
(c) V_A < V_B, Z_A > Z_B
(d) V_A > V_B, Z_A < Z_B

Q. Figure shows intensity wavelength relation of X-rays in two different tubes A and B operating at

V_{A}

and $V_{B}$ having targets of atomic number $Z_{A}$ and $Z_{B}$ then

Q. Given curve shows the intensity wavelength relation of

X -

rays coming from two different Coolidge tubes

A

and

B

. The dark curve represents the relation for the tube

A

in which potential difference between the target and the filament is

V_{A}

and the atomic number of the target material is

Z_{A}

. Similarly dotted curve is for tube

B

. Respective quantities are

V_{B}

and

Z_{B}

foe the tube

B

. The,

Q. Two element A and B with atomic numbers

Z_{A}

and

Z_{B}

are used to produce characteristic X-rays with frequencies

v_{A}

and

v_{B}

respectively. If

Z_{A} : Z_{B} = 1 : 2

, then

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will be

A mole of $C C l_{4}$ vapours at $77^{\circ} C$ occupies a volume of 36.0 L.If vanderwaals constants are $a = 20.39 L^{2} a t m m o l^{- 2}$ and $b = 0.1383 L m o l^{- 1}$ calculate the compressibility factor Z under (a) low pressure region and (b) high pressure region.
The values of compressibility factor Za and Zb respectively are:

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The correct option is B VA>VB,ZA<ZBλmin of A<λmin of Bso VA>VBalso λKα,A>λKα,Bso ZB>ZA

The correct option is B $V_{A} > V_{B}, Z_{A} < Z_{B}$
$λ_{m i n} o f A < λ_{m i n} o f B$
so $V_{A} > V_{B}$
also $λ_{K_{α, A}} > λ_{K_{α, B}}$
so $Z_{B} > Z_{A}$