Atomic number of anti cathode material in an x-ray tube is 41 wavelength of care for x-ray produced
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Answered by
135
Explanation:
Answer
For k_ak
a
x-ray line
\dfrac{1}{\lambda_o}=R(z-1)^2\left[1-\dfrac{1}{2^2}\right]
λ
o
1
=R(z−1)
2
[1−
2
2
1
] z=41z=41
=R[41-1]^2\left[1-\dfrac{1}{4}\right]=R[41−1]
2
[1−
4
1
] R=1.097\times 10^7R=1.097×10
7
\dfrac{1}{\lambda}=1.097\times 10^7m[40]^2\left[1-\dfrac{1}{4}\right]
λ
1
=1.097×10
7
m[40]
2
[1−
4
1
]
Solving =0.76 \mathring{A}=0.76\times 10^{10}m=0.76
A
˚
=0.76×10
10
m
\lambda \therefore 0.76 \mathring{A}λ∴0.76
A
˚
.
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