1.
"f = \\dfrac{c}{\\lambda} = \\dfrac{3 \\cdot 10^{8} \\dfrac{m}{s}}{442 \\cdot 10^{-9} m} = 6.78 \\cdot 10^{14} Hz."
Since,
"f>f_0"the electrons will be ejected from the surface of the sodium.
2.
"v = \\sqrt{\\dfrac{2h(f - f_0)}{m}}"
"v = \\sqrt{\\dfrac{2 \\cdot 6.626 \\cdot 10^{-34} \\cdot ( 6.78 \\cdot 10^{14} - 5.71 \\cdot 10^{14} )}{9.1 \\cdot 10^{-31} }} = 3.95 \\cdot 10^5 \\dfrac{m}{s}."
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