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In a discharge tube, which contains argon at low pressure, V potential difference is applied between two of its electrodes. Ionisation energy of argon atom is 15.6 eV. Separation between electrodes is 4.0 xx 10^(-2) m and average distance that electron travels between two successive collisions with argon atoms is 8 xx 10^(-5)m. Estimate the minimum value of V such that collision of electron may cause ionisation of argon atoms. |
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Answer» Solution :Electric field between the plates can be WRITTEN as follows : E=V/d , where `d(=4.0xx10^(-2)m)` is separationbetween the plates. If `lambda(=8xx10^(-5)m)` is distance between two successive collisions then ENERGY ACQUIRED by the electron can be written as follows : Energy acquired by electron before collision `=eElambda=eVlambda//d` This energy must be equal to 15.6 eV to cause ionisation of argon atom. `(eVlambda)/d=15.6xxe` `V=(15.6xxd)/lambda=(15.6xx4xx10^(-2))/(8xx10^5)` `=7.8xx10^3` V = 7.8 KV |
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