The work function of sodium metal is 4.41 $\times$ 10–19 J. If photons of wavelength 300 nm are incident on the metal, the kinetic energy of the ejected electrons will be (h = 6.63 $\times$ 10–34 J s; c = 3 $\times$ 108 m/s) ________ × 10–21 J.
Answer (integer)
222
Solution
E = W + KE<sub>max</sub>
<br><br>$\Rightarrow$ KE<sub>max</sub> = E - W
<br><br>= ${{hc} \over \lambda }$ - 4.41 $\times$ 10<sup>–19</sup>
<br><br>= $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {300 \times {{10}^{ - 9}}}}$$ - 4.41 $\times$ 10<sup>–19</sup>
<br><br>= 2.22 × 10<sup>–19</sup> J
<br><br>= 222 × 10<sup>–21</sup> J
About this question
Subject: Chemistry · Chapter: Atomic Structure · Topic: Atomic Spectra
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