Home
Class 11
CHEMISTRY
The characteristic X-rays for the lines ...

The characteristic X-rays for the lines of `K_(a)` series in element X and Y are `9.87 Å and 14.6 Å` respectively .If Moseley's equation `sqrt(v) = 4.9 xx 10^(7) (Z - 0.75)` is followed:
The atomic number of X is

A

8

B

10

C

12

D

16

Text Solution

AI Generated Solution

The correct Answer is:
To find the atomic number of element X using the given information, we will follow these steps: ### Step 1: Understand the problem We are given the characteristic X-ray wavelengths for the Kα series of two elements, X and Y, which are 9.87 Å and 14.6 Å, respectively. We need to find the atomic number (Z) of element X using Moseley's equation. ### Step 2: Convert wavelength to frequency Moseley's equation relates the frequency (ν) of the X-ray to the atomic number (Z). We first need to convert the wavelength (λ) of element X into frequency using the equation: \[ \nu = \frac{c}{\lambda} \] where \( c \) is the speed of light (approximately \( 3 \times 10^8 \) m/s). For element X: \[ \lambda = 9.87 \, \text{Å} = 9.87 \times 10^{-10} \, \text{m} \] Now, substituting the values: \[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{9.87 \times 10^{-10} \, \text{m}} \] ### Step 3: Calculate the frequency Calculating the frequency: \[ \nu = \frac{3 \times 10^8}{9.87 \times 10^{-10}} \approx 3.03 \times 10^{18} \, \text{s}^{-1} \] ### Step 4: Apply Moseley's equation Moseley's equation is given by: \[ \sqrt{\nu} = 4.9 \times 10^7 (Z - 0.75) \] First, we need to calculate \( \sqrt{\nu} \): \[ \sqrt{\nu} = \sqrt{3.03 \times 10^{18}} \approx 5.50 \times 10^9 \, \text{s}^{-1} \] ### Step 5: Substitute into Moseley's equation Now we substitute \( \sqrt{\nu} \) into Moseley's equation: \[ 5.50 \times 10^9 = 4.9 \times 10^7 (Z - 0.75) \] ### Step 6: Solve for Z To isolate Z, we first divide both sides by \( 4.9 \times 10^7 \): \[ Z - 0.75 = \frac{5.50 \times 10^9}{4.9 \times 10^7} \] Calculating the right-hand side: \[ Z - 0.75 \approx 112.24 \] Now, adding 0.75 to both sides: \[ Z \approx 112.24 + 0.75 \] \[ Z \approx 113.00 \] ### Step 7: Conclusion The atomic number of element X is approximately 113.

To find the atomic number of element X using the given information, we will follow these steps: ### Step 1: Understand the problem We are given the characteristic X-ray wavelengths for the Kα series of two elements, X and Y, which are 9.87 Å and 14.6 Å, respectively. We need to find the atomic number (Z) of element X using Moseley's equation. ### Step 2: Convert wavelength to frequency Moseley's equation relates the frequency (ν) of the X-ray to the atomic number (Z). We first need to convert the wavelength (λ) of element X into frequency using the equation: \[ \nu = \frac{c}{\lambda} \] ...
Promotional Banner

Topper's Solved these Questions

  • ATOMIC STRUCTURE

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Multiple Correct|45 Videos
  • ATOMIC STRUCTURE

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Single Correct|127 Videos
  • ATOMIC STRUCTURE

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Subjective)|52 Videos
  • APPENDIX - INORGANIC VOLUME 1

    CENGAGE CHEMISTRY ENGLISH|Exercise chapter-7 Single correct answer|1 Videos
  • CHEMICAL BONDING AND MOLECULAR STRUCTURE

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives Subjective|15 Videos

Similar Questions

Explore conceptually related problems

The characteristic X-rays for the lines of K_(a) series in element X and Y are 9.87 Å and 14.6 Å respectively .If Moseley's equation sqrt(v) = 4.9 xx 10^(7) (Z - 0.75) is followed: The atomic number of Y is

Elements X, Y and Z have atomic numbers 6, 9 and 12 respectively. Which one : forms a cation

Elements X, Y and Z have atomic numbers 6, 9 and 12 respectively. Which one : forms an anion

The k_alpha X-rays of aluminium (Z = 13 ) and zinc ( Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley\'s law sqrt v = a(Z - b) to find the wavelength of the K_alpha X-ray of iron (Z = 26).

The k_alpha X-rays of aluminium (Z = 13 ) and zinc ( Z = 30) have wavelengths 887 pm and 146 pm respectively. Use Moseley\'s law sqrt v = a(Z - b) to find the wavelength of the K_alpha X-ray of iron (Z = 26).

The equation sqrt(4x+9)-sqrt(11x+1)=sqrt(7x+4) has

Moseley's law for characteristic X-rays is sqrt v = a(Z-b) . In this,

Two lines of regressions are represented by 4x + 10y = 9 and 6x + 3y = 4 . Find the line of regression y on x.

Elements X, Y and Z have atomic numbers 6, 9 and 12 respectively. Which one : has four electrons in the valency shell ?

In Moseley's law sqrt(v)=a(z-b), the volue of the screening constant for K-series and L-series of X-rays are respectively

CENGAGE CHEMISTRY ENGLISH-ATOMIC STRUCTURE-Exercises Linked Comprehension
  1. A hydrogen like atom (atomic number Z) is in a higher excited satte ...

    Text Solution

    |

  2. A hydrogen like species (atomic number Z) is present in a higher excit...

    Text Solution

    |

  3. The characteristic X-rays for the lines of K(a) series in element X a...

    Text Solution

    |

  4. The characteristic X-rays for the lines of K(a) series in element X a...

    Text Solution

    |

  5. Werner Heisenberg considered the limits of how precisely we can measur...

    Text Solution

    |

  6. It is impossible to determine simultaneously the position of velocity...

    Text Solution

    |

  7. The seqence of filling electgron in sub-shells of element with few ex...

    Text Solution

    |

  8. If Hund's rule is not obeyed by some element given below then whi...

    Text Solution

    |

  9. The sequence of filling electron in sub-shells of element with few ex...

    Text Solution

    |

  10. The sequence of filling electron in sub-shells of element with few ex...

    Text Solution

    |

  11. The sepence of filling electgron in sub-shells of element with few ex...

    Text Solution

    |

  12. The only element in the hydrogen atom resides under ordinary conditi...

    Text Solution

    |

  13. The only element in the hydrogen atom resides under ordinary conditi...

    Text Solution

    |

  14. The only element in the hydrogen atom resides under ordinary conditi...

    Text Solution

    |

  15. The only element in the hydrogen atom resides under ordinary conditi...

    Text Solution

    |

  16. The shape of orbitals are related to the ratio of principal quantum ...

    Text Solution

    |

  17. The shape of orbitals are related to the ratio of principal quantum ...

    Text Solution

    |

  18. The shape of orbitals are related to the ratio of principal quantum ...

    Text Solution

    |

  19. The shape of orbitals are related to the ratio of principal quantum ...

    Text Solution

    |

  20. The shape of orbitals are related to the ratio of principal quantum ...

    Text Solution

    |