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Elijah Hammonds - Wavelength Frequency and Energy - Chemistry ... - Free Printable

Elijah Hammonds - Wavelength Frequency and Energy - Chemistry ...

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Show Answer Key & Explanations Step-by-step solution for: Elijah Hammonds - Wavelength Frequency and Energy - Chemistry ...
1. λ = 3.00 × 10⁸ m/s / 5.76 × 10¹⁴ Hz = 5.21 × 10⁻⁷ m
2. f = 3.00 × 10⁸ m/s / 6.50 × 10⁻⁷ m = 4.62 × 10¹⁴ Hz
3. E = (6.626 × 10⁻³⁴ J·s) × (5.76 × 10¹⁴ Hz) = 3.82 × 10⁻¹⁹ J
4. λ = 3.00 × 10⁸ m/s / 1.20 × 10¹⁵ Hz = 2.50 × 10⁻⁷ m
5. f = 3.00 × 10⁸ m/s / 4.50 × 10⁻⁷ m = 6.67 × 10¹⁴ Hz
6. E = (6.626 × 10⁻³⁴ J·s) × (1.20 × 10¹⁵ Hz) = 7.95 × 10⁻¹⁹ J
7. λ = 3.00 × 10⁸ m/s / 2.40 × 10¹⁴ Hz = 1.25 × 10⁻⁶ m
8. f = 3.00 × 10⁸ m/s / 1.25 × 10⁻⁶ m = 2.40 × 10¹⁴ Hz
9. E = (6.626 × 10⁻³⁴ J·s) × (2.40 × 10¹⁴ Hz) = 1.59 × 10⁻¹⁹ J
10. λ = 3.00 × 10⁸ m/s / 1.50 × 10¹⁵ Hz = 2.00 × 10⁻⁷ m
11. f = 3.00 × 10⁸ m/s / 2.00 × 10⁻⁷ m = 1.50 × 10¹⁵ Hz
12. E = (6.626 × 10⁻³⁴ J·s) × (1.50 × 10¹⁵ Hz) = 9.94 × 10⁻¹⁹ J
13. λ = 3.00 × 10⁸ m/s / 3.00 × 10¹⁴ Hz = 1.00 × 10⁻⁶ m
14. f = 3.00 × 10⁸ m/s / 1.00 × 10⁻⁶ m = 3.00 × 10¹⁴ Hz
15. E = (6.626 × 10⁻³⁴ J·s) × (3.00 × 10¹⁴ Hz) = 1.99 × 10⁻¹⁹ J
16. λ = 3.00 × 10⁸ m/s / 6.00 × 10¹⁴ Hz = 5.00 × 10⁻⁷ m
17. f = 3.00 × 10⁸ m/s / 5.00 × 10⁻⁷ m = 6.00 × 10¹⁴ Hz
18. E = (6.626 × 10⁻³⁴ J·s) × (6.00 × 10¹⁴ Hz) = 3.98 × 10⁻¹⁹ J
19. λ = 3.00 × 10⁸ m/s / 1.00 × 10¹⁵ Hz = 3.00 × 10⁻⁷ m
20. f = 3.00 × 10⁸ m/s / 3.00 × 10⁻⁷ m = 1.00 × 10¹⁵ Hz
21. E = (6.626 × 10⁻³⁴ J·s) × (1.00 × 10¹⁵ Hz) = 6.63 × 10⁻¹⁹ J
22. λ = 3.00 × 10⁸ m/s / 2.00 × 10¹⁵ Hz = 1.50 × 10⁻⁷ m
23. f = 3.00 × 10⁸ m/s / 1.50 × 10⁻⁷ m = 2.00 × 10¹⁵ Hz
24. E = (6.626 × 10⁻³⁴ J·s) × (2.00 × 10¹⁵ Hz) = 1.33 × 10⁻¹⁸ J
Parent Tip: Review the logic above to help your child master the concept of wavelength and frequency worksheet.
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