Step-by-step explanation:
Given that,
The frequency of electromagnetic spectrum is
![2.73* 10^(16)\ Hz](https://img.qammunity.org/2021/formulas/chemistry/college/tic2cwuhta15g0h1n9jrgp4j1806bqpf3h.png)
(A) Let the wavelength of this radiation is
. We know that,
![c=f\lambda\\\\\lambda=(c)/(f)\\\\\lambda=(3* 10^8)/(2.73* 10^(16))\\\\\lambda=1.09* 10^(-8)\ m](https://img.qammunity.org/2021/formulas/chemistry/college/rbnwced40198e8dier9y6a0tf3zvciyq87.png)
So, the wavelength of this radiation is
.
(B) Let E is the energy associated with this radiation. Energy of an electromagnetic radiation is given by :
![E=hf](https://img.qammunity.org/2021/formulas/physics/college/yqhzwd2c5hhi6554ecj68o32z9h0zzkgs6.png)
h is Planck's constant
![E=6.63* 10^(-34)* 2.73* 10^(16)\\\\E=1.8* 10^(-17)\ J](https://img.qammunity.org/2021/formulas/chemistry/college/b3y23qiz21n9ets5odfpl10ll50xv1s6uq.png)
1 kcal = 4184 J
It means,
![1.8* 10^(-17)\ J=(1)/(4184)* 1.8* 10^(-17)\\\\=4.3* 10^(-21)\ \text{kcal}](https://img.qammunity.org/2021/formulas/chemistry/college/tut4rxb0jxmtpta6d9lfwndmhlskermhos.png)
Hence, this is the required solution.